Arithmetic Progression

Show that the progression 11, 13, 15, 17, 19,........is an A.P. Write down its :

(i) first term

(ii) common difference

(iii) 20th term.

Answer

Since, 13 - 11 = 2, 15 - 13 = 2, 17 - 15 = 2 and 19 - 17 = 2.

Hence, the series is an A.P. with common difference = 2.

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₂₀ = 11 + (20 - 1)2

= 11 + (19)2

= 11 + 38

= 49.

Hence, a = 11, d = 2 and a₂₀ = 49.

Show that the progression 13, 20, 27, 34, ........ is an A.P. Write down its :

(i) first term

(ii) common difference

(iii) 20th term.

Answer

Since, 20 - 13 = 7, 27 - 20 = 7, and 34 - 27 = 7.

Hence, the series is an A.P. with common difference = 7.

We know that,

n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₂₀ = 13 + (20 - 1)7

= 13 + (19) × 7

= 13 + 133

= 146.

Hence, a = 13, d = 7 and a₂₀ = 146.

Show that the progression 10, 6, 2, -2, -6,........is an A.P. Write down its

(i) first term

(ii) common difference

(iii) 10th term.

Answer

Since, 6 - 10 = -4, 2 - 6 = -4, -2 - 2 = -4 and -6 - (-2) = -4.

Hence, the series is an A.P. with common difference = -4.

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₁₀ = 10 + (10 - 1).(-4)

= 10 + (9)-4

= 10 - 36

= -26.

Hence, a = 10, d = -4 and a₂₀ = -26.

Show that the progression 11, 5, -1, -7, -13,........is an A.P. Write down its

(i) first term

(ii) common difference

(iii) 12th term.

Answer

Since, 5 - 11 = -6, -1 - 5 = -6, -7 - (-1) = -6 and -13 - (-7) = -6.

Hence, the series is an A.P. with common difference = -6.

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₁₂ = 11 + (12 - 1).(-6)

= 11 + 11(-6)

= 11 - 66

= -55.

Hence, a = 11, d = -6 and a₁₂ = -55.

Show that the progression 6, $7\dfrac{3}{4}, 9\dfrac{1}{2}, 11\dfrac{1}{4}$,........is an A.P. Write down its

(i) first term

(ii) common difference

(iii) 17th term.

Answer

$7\dfrac{3}{4} = \dfrac{31}{4} = 7.75 \\[1em] 9\dfrac{1}{2} = \dfrac{19}{2} = 9.5 \\[1em] 11\dfrac{1}{4} = \dfrac{45}{4} = 11.25 \\[1em]$

Since, 7.75 - 6 = 1.75, 9.5 - 7.75= 1.75 and 11.25 - 9.5 = 1.75.

Hence, the series is an A.P. with common difference = 1.75.

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₁₇ = 6 + (17 - 1)1.75

= 6 + (16)1.75

= 6 + 28

= 34.

Hence, a = 6, d = $\dfrac{7}{4}$ and a₁₇ = 34.

Show that the progression 6, $5\dfrac{1}{2}, 5, 4\dfrac{1}{2}, 4$,........is an A.P. Write down its

(i) first term

(ii) common difference

(iii) 9th term.

Answer

Series :

$\Rightarrow 6, 5\dfrac{1}{2}, 5, 4\dfrac{1}{2}, 4$

⇒ 6, 5.5, 5, 4.5, 4

Since, 5.5 - 6 = -0.5, 5 - 5.5 = -0.5, 4.5 - 5 = -0.5 and 4.0 - 4.5 = -0.5.

Hence, the series is an A.P. with common difference (d) = -0.5

We know that n^th term of an A.P. is given by,

⇒ a_n = a + (n - 1)d, where a is the first term.

⇒ a₉ = 6 + (9 - 1)(-0.5)

= 6 + 8(-0.5)

= 6 - 4

= 2.

Hence, a = 6, d = -0.5 and a₉ = 2.

Find the nth term of the A.P. 5, 11, 17, 23,....... Using it, find its

(i) 11th term

(ii) 21st term.

Answer

Given,

a = 5

d = 11 - 5 = 6

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

a_n = 5 + (n - 1)6

= 5 + 6n - 6

= 6n - 1.

(i) 11th term:

⇒ a₁₁ = 5 + (11 - 1)6

= 5 + 10(6)

= 5 + 60

= 65.

(ii) 21st term

⇒ a₂₁ = 5 + (21 - 1)6

= 5 + (20)6

= 5 + 120

= 125.

Hence, nth term = 6n - 1, 11th term = 65 and 21st term = 125.

Find the nth term of the A.P. 13, 7, 1, -5, -11, ....... Using it, find its

(i) 9th term

(ii) 16th term.

Answer

Given,

a = 13

d = 7 - 13 = -6

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

a_n = 13 + (n - 1)(-6)

= 13 - 6n + 6

= 19 - 6n.

(i) 9th term

⇒ a₉ = 13 + (9 - 1).(-6)

= 13 + 8(-6)

= 13 - 48

= -35.

(ii) 16th term

⇒ a₁₆ = 13 + (16 - 1).(-6)

= 13 + (15).(-6)

= 13 - 90

= -77.

Hence, nth term = 19 - 6n, 9th term = -35 and 16th term = -77.

How many terms are there in the A.P. 6, 10, 14, 18, ...., 174 ?

Answer

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given,

⇒ a = 6

⇒ d = 10 - 6 = 4

⇒ a_n = 174

⇒ 174 = 6 + (n - 1)4

⇒ 174 - 6 = (n - 1)4

⇒ 168 = (n - 1)4

⇒ $\dfrac{168}{4}$ = n - 1

⇒ 42 = n - 1

⇒ n = 42 + 1

⇒ n = 43.

Hence, there are 43 terms in the A.P.

How many terms are there in the A.P. 41, 38, 35 ,...., -1 ?

Answer

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given,

a = 41

d = 38 - 41 = -3

Let nth term be -1.

⇒ a_n = -1

⇒ a + (n - 1)d = -1

⇒ -1 = 41 + (n - 1)(-3)

⇒ - 1 - 41 = (n - 1)(-3)

⇒ -42 = (n - 1)(-3)

⇒ $\dfrac{-42}{-3}$ = n - 1

⇒ 14 = n - 1

⇒ n = 14 + 1

⇒ n = 15.

Hence, there are 15 terms in the A.P.

Which term of A.P. 3, 8, 13, 18, 23,...., is 98?

Answer

In the A.P. 3, 8, 13, 18, 23,...., a = 3 and d = 8 - 3 = 5.

Let n^th term be 98.

∴ a_n = 98

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

⇒ 98 = 3 + (n - 1)5

⇒ 98 - 3 = (n - 1)5

⇒ 95 = (n - 1)5

⇒ (n - 1) = $\dfrac{95}{5}$

⇒ n - 1 = 19

⇒ n = 19 + 1

⇒ n = 20.

Hence, 20th term of A.P. is 98.

Which term of A.P. 72, 68, 64, 60, ...., is 0 ?

Answer

In the A.P. 72, 68, 64, 60,...., a = 72 and d = 68 - 72 = -4.

Let n^th term be 0.

∴ a_n = 0

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

⇒ 0 = 72 + (n - 1)(-4)

⇒ -72 = (n - 1)(-4)

⇒ n - 1 = $\dfrac{-72}{-4}$

⇒ n - 1 = 18

⇒ n = 18 + 1

⇒ n = 19.

Hence, 19th term of A.P. is 0.

Which term of A.P. 1, $1\dfrac{1}{6}, 1\dfrac{1}{3}$,....,is 3?

Answer

Series :

$\Rightarrow 1, 1\dfrac{1}{6}, 1\dfrac{1}{3}, ......$

$\Rightarrow 1, \dfrac{7}{6}, \dfrac{4}{3}, ......$

In the A.P. 1, $1\dfrac{1}{6}, 1\dfrac{1}{3}$,...., a = 1 and d = $\dfrac{7}{6} - 1 = \dfrac{1}{6}$

Let n^th term be 3.

∴ a_n = 3

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

$\Rightarrow 3 = 1 + (n - 1)\dfrac{1}{6} \\[1em] \Rightarrow 3 = 1 + (n - 1)\dfrac{1}{6} \\[1em] \Rightarrow 3 - 1 = (n - 1)\dfrac{1}{6} \\[1em] \Rightarrow 2 = (n - 1)\dfrac{1}{6} \\[1em] \Rightarrow 2 \times 6 = (n - 1) \\[1em] \Rightarrow 12 = (n - 1) \\[1em] \Rightarrow n = 12 + 1 \\[1em] \Rightarrow n = 13.$

Hence, 13th term of A.P. is 3.

The 4th and 10th terms of A.P. are 13 and 25 respectively. Find its (i) first term (ii) common difference (iii) 17th term.

Answer

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given, 4th term is 13.

∴ a₄ = a + (4 - 1)d

⇒ 13 = a + 3d

⇒ a + 3d = 13 .......(1)

Given, 10th term is 25

∴ a₁₀ = a + (10 - 1)d

⇒ 25 = a + 9d

⇒ a + 9d = 25 ......(2)

Subtracting equation (1) from (2) we get,

⇒ a + 9d - (a + 3d) = 25 - 13

⇒ a + 9d - a - 3d = 12

⇒ 6d = 12

⇒ d = $\dfrac{12}{6}$

⇒ d = 2.

Substituting value of d in (1) we get,

⇒ a + 3(2) = 13

⇒ a + 6 = 13

⇒ a = 13 - 6

⇒ a = 7.

∴ a₁₇ = 7 + (17 - 1)2

= 7 + (16)2

= 7 + 32

= 39.

Hence, a = 7, d = 2 and a₁₇ = 39.

The 7th term of an A.P. is -4 and its 13th term is -16. Find its (i) first term (ii) common difference (iii) nth term.

Answer

Let first term be a and common difference be d.

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given, 7th term is -4.

∴ a₇ = a + (7 - 1)d

⇒ -4 = a + 6d

⇒ a + 6d = -4 ....(1)

Given,

13th term is -16

∴ a₁₃ = a + (13 - 1)d

⇒ -16 = a + 12d

⇒ a + 12d = -16 ....(2)

Subtracting (1) from (2) we get,

⇒ a + 12d - (a + 6d) = -16 - (-4)

⇒ a + 12d - a - 6d = -16 + 4

⇒ 6d = -12

⇒ d = $\dfrac{-12}{6}$

⇒ d = -2.

Substituting value of d in (1) we get,

⇒ a + 6(-2) = -4

⇒ a - 12 = -4

⇒ a = -4 + 12

⇒ a = 8.

∴ a_n = 8 + (n - 1)(-2)

= 8 - 2n + 2

= 10 - 2n.

Hence, a = 8, d = -2 and a_n = 10 - 2n.

Find the 8th term from end of the A.P. 7, 10, 13,....., 184.

Answer

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given,

a = 7

d = 10 - 7 = 3

⇒ a_n = 184

⇒ a_n = 7 + (n - 1)3

⇒ 184 = 7 + (n - 1)3

⇒ 184 - 7 = (n -1)3

⇒ 177 = (n - 1)3

⇒ $\dfrac{177}{3}$ = (n - 1)

⇒ (n - 1) = 59

⇒ n = 59 + 1

⇒ n = 60.

There are 60 terms in the A.P.

The 8th term from end of the A.P. is 53rd (60 - 7) term from the starting.

⇒ a₅₃ = 7 + (53 - 1)3

= 7 + (52)3

= 7 + 156

= 163.

Hence, the 8th term from end of the A.P is 163.

Find the 6th term from end of the A.P. 17, 14, 11,....., (-40).

Answer

We know that,

n^th term of an A.P. is given by,

a_n = a + (n - 1)d

Given,

a = 17

d = 14 - 17 = -3

Let no. of terms be n.

⇒ a_n = -40

By formula,

⇒ a_n = a + (n - 1)d

⇒ -40 = 17 + (n - 1)(-3)

⇒ -40 - 17 = (n - 1)(-3)

⇒ -57 = (n - 1)(-3)

⇒ $\dfrac{-57}{-3}$ = n - 1

⇒ 19 = n - 1

⇒ n = 19 + 1

⇒ n = 20.

There are 20 terms in the A.P.

The 8th term from end of the A.P. = 20 - 5 = 15

⇒ a₁₅ = 17 + (15 - 1)-3

= 17 + (14)-3

= 17 - 42

= -25

Hence, the 6th term from end of the A.P is -25.

Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in A.P.

Answer

Since, (5x + 2), (4x - 1) and (x + 2) are in A.P.

Hence, difference between consecutive terms are equal.

∴ (4x - 1) - (5x + 2) = (x + 2) - (4x - 1)

⇒ 4x - 1 - 5x - 2 = x + 2 - 4x + 1

⇒ -3 - x = 3 - 3x

⇒ -3 - 3 = -3x + x

⇒ -6 = -2x

⇒ x = $\dfrac{-6}{-2}$

⇒ x = 3.

Hence, the value of x = 3.

If (k - 3), (2k + 1) and (4k + 3) are three consecutive terms of an A.P., find the value of k.

Answer

Since, (k - 3), (2k + 1) and (4k + 3) are in A.P.

Hence, difference between consecutive terms are equal.

∴ 2k + 1 - (k - 3) = 4k + 3 - (2k + 1)

⇒ 2k + 1 - k + 3 = 4k + 3 - 2k - 1

⇒ k + 4 = 2k + 2

⇒ 4 - 2 = 2k - k

⇒ k = 2.

Hence, the value of k = 2.

Find three numbers in A.P., whose sum is 15 and the product is 80.

Answer

Let the three numbers in the Arithmetic Progression (A.P.) be a - d, a, a + d.

Given,

The sum of the three numbers is 15.

⇒ a - d + a + a + d = 15

⇒ 3a = 15

⇒ a = $\dfrac{15}{3}$

⇒ a = 5.

The middle term of the A.P. is 5.

Given,

The product of the three numbers is 80.

⇒ (a - d)(a)(a + d) = 80

⇒ (5 - d)(5)(5 + d) = 80

⇒ (5 - d)(5 + d) = $\dfrac{80}{5}$

⇒ (5 - d)(5 + d) = 16

⇒ 5² - d² = 16

⇒ 25 - d² = 16

⇒ d² = 25 - 16

⇒ d² = 9

⇒ d = $\sqrt{9}$

⇒ d = 3 or d = -3.

Case 1 : If a = 5 and d = 3

⇒ a - d = 5 - 3 = 2

⇒ a = 5

⇒ a + d = 5 + 3 = 8.

Case 2 : If a = 5 and d = -3

⇒ a - d = 5 - (-3) = 8

⇒ a = 5

⇒ a + d = 5 + (-3) = 2.

Hence, the numbers are 2, 5, 8.

The angles of quadrilateral are in A.P., whose common difference is 10°. Find the angles.

Answer

Let the four angles of the quadrilateral be in A.P. with common difference 10° be:

a, a + 10°, a + 20°, a + 30°

The sum of the interior angles of any quadrilateral is always 360°.

⇒ a + a + 10° + a + 20° + a + 30° = 360°

⇒ 4a + 60° = 360°

⇒ 4a = 360° - 60°

⇒ 4a = 300°

⇒ a = $\dfrac{300°}{4}$

⇒ a = 75°

⇒ a + 10° = 75° + 10° = 85°

⇒ a + 20° = 75° + 20° = 95°

⇒ a + 30° = 75° + 30° = 105°.

Hence, the angles of quadrilateral are 75°, 85°, 95°, 105°.

The angles of triangle are in A.P. whose common difference is 20°. Find angles.

Answer

Let the three angles of triangle in A.P. be represented as : a - d, a, a + d.

The sum of the interior angles of any triangle is always 180°.

⇒ a - d + a + a + d = 180°

⇒ 3a = 180

⇒ a = $\dfrac{180}{3}$

⇒ a = 60°.

Given,

The common difference is 20°

⇒ a - 20° = 60° - 20° = 40°

⇒ a + 20° = 60° + 20° = 80°.

Hence, the angles of triangle are 40°, 60°, 80°.

Find the sum :

2 + 7 + 12 + 17 + ..... to 19 terms.

Answer

The above sequence is an A.P. with common difference = 12 - 7 = 5.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₁₉ = $\dfrac{19}{2}$ [2(2) + (19 - 1)5]

= 9.5 [4 + (18)5]

= 9.5 [4 + 90]

= 9.5 (94)

= 893.

Hence, S₁₉ = 893.

Find the sum :

9 + 7 + 5 + 3 +......to 14 terms.

Answer

The above sequence is an A.P. with common difference = 5 - 7 = -2.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₁₄ = $\dfrac{14}{2}$ [2(9) + (14 - 1)(-2)]

= 7[18 + (13)(-2)]

= 7[18 - 26]

= 7.(-8)

= -56.

Hence, S₁₄ = -56.

Find the sum :

(-11) + (-7) + (-3) + 1 +......to 12 terms.

Answer

The above sequence is an A.P. with common difference = -7 - (-11) = 4.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₁₂ = $\dfrac{12}{2}$ [2(-11) + (12 - 1)4]

= 6.[-22 + (11)4]

= 6.[-22 + 44]

= 6.(22)

= 132.

Hence, S₁₂ = 132.

Find the sum :

$\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}$, ......to 11 terms.

Answer

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Given,

a = $\dfrac{1}{15}$

d = $\dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5 - 4}{60} = \dfrac{1}{60}$

n = 11

$\Rightarrow S_{11} = \dfrac{11}{2} \Big[2\Big(\dfrac{1}{15}\Big) + (11 - 1)\Big(\dfrac{1}{60}\Big)\Big] \\[1em] = \dfrac{11}{2} \Big[\Big(\dfrac{2}{15}\Big) + \Big(\dfrac{10}{60}\Big)\Big] \\[1em] = \dfrac{11}{2} \Big[\dfrac{2 \times 2 + 1 \times 5 }{30}\Big] \\[1em] = \dfrac{11}{2} \Big[\dfrac{4 + 5}{30}\Big] \\[1em] = \dfrac{11}{2} \Big[\dfrac{9}{30}\Big] \\[1em] = \dfrac{11}{2} \Big[\dfrac{3}{10}\Big] \\[1em] = \dfrac{33}{20}.$

Hence, S₁₁ = $\dfrac{33}{20}$.

Find the sum :

0.6 + 1.7 + 2.8 +..... to 100 terms.

Answer

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Given,

a = 0.6

d = 1.7 - 0.6 = 1.1

n = 100

⇒ S₁₀₀ = $\dfrac{100}{2}$ [2(0.6) + (100 - 1)1.1]

= 50 [1.2 + (99)1.1]

= 50 [1.2 + 108.9]

= 50 (110.1)

= 5505.

Hence, S₁₀₀ = 5505.

Find the sum :

7 + $10\dfrac{1}{2}$ + 14 +......+ 84.

Answer

Series :

7 + 10.5 + 14 + ....... + 84

The above series is an A.P. with common difference = 3.5

Let 84 be nth term in the A.P.

We know that,

∴ a_n = a + (n - 1)d

⇒ 84 = 7 + (n - 1) × 3.5

⇒ 84 - 7 = (n - 1) × 3.5

⇒ 77 = (n - 1) × 3.5

⇒ $\dfrac{77}{3.5}$ = (n - 1)

⇒ 22 = (n - 1)

⇒ n = 22 + 1

⇒ n = 23.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ (a + l)

⇒ S₂₃ = $\dfrac{23}{2}$ (7 + 84)

= $\dfrac{23}{2}$ (91)

= $\dfrac{2093}{2} = 1046\dfrac{1}{2}$

Hence, S₂₃ = $1046\dfrac{1}{2}$.

Find the sum :

32 + 30 + 28 +........+ 10.

Answer

Let total number of terms in the A.P. be n.

We know that,

∴ a_n = a + (n - 1)d

⇒ 10 = 32 + (n - 1).(-2)

⇒ 10 - 32 = (n - 1).(-2)

⇒ -22 = (n - 1).(-2)

⇒ $\dfrac{-22}{-2}$ = (n - 1)

⇒ 11 = n - 1

⇒ n = 11 + 1

⇒ n = 12.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ (a + l)

⇒ S₁₂ = $\dfrac{12}{2}$ (32 + 10)

= 6 × 42

= 252.

Hence, S₁₂ = 252.

Find the sum :

(-5) + (-8) + (-11) +....... + (-62).

Answer

Let total number of terms in an A.P. be n.

We know that,

∴ a_n = a + (n - 1)d

Given,

a = -5

d = -8 - (-5) = -3

⇒ a_n = -5 + (n - 1)-3

⇒ -62 = -5 + (n - 1)-3

⇒ -62 - (-5) = (n - 1)-3

⇒ -62 - (-5) = (n - 1)-3

⇒ -57 = (n - 1)-3

⇒ $\dfrac{-57}{-3}$ = (n - 1)

⇒ 19 = (n - 1)

⇒ n = 19 + 1

⇒ n = 20

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ (a + l)

⇒ S₂₀ = $\dfrac{20}{2}$ [-5 + (-62)]

= 10 [-5 + (-62)]

= 10(-67)

= -670.

Hence, S₂₀ = -670.

Find the sum of all 2-digit natural numbers divisible by 5.

Answer

The 2-digit natural numbers divisible by 5 form an A.P., with common difference = 5.

10, 15, 20,.....,95

Let total number of terms in an A.P. be n.

We know that,

∴ a_n = a + (n - 1)d

⇒ 95 = 10 + (n - 1)5

⇒ 95 - 10 = (n - 1)5

⇒ 85 = (n - 1)5

⇒ $\dfrac{85}{5}$ = (n - 1)

⇒ 17 = (n - 1)

⇒ n = 17 + 1

⇒ n = 18.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ (a + l)

⇒ S₁₈ = $\dfrac{18}{2} \times (10 + 95)$

= 9(105)

= 945.

Hence, S₁₈ = 945.

Find the sum of all even numbers between 10 and 100.

Answer

Even numbers between 10 and 100 are :

12, 14, 16, ....., 98

Let total number of terms in A.P. be n.

Given,

a = 12

d = 14 - 12 = 2

We know that,

∴ a_n = a + (n - 1)d

⇒ 98 = 12 + (n - 1)2

⇒ 98 - 12 = (n - 1)2

⇒ 86 = (n - 1)2

⇒ $\dfrac{86}{2}$ = (n - 1)

⇒ n - 1 = 43

⇒ n = 43 + 1

⇒ n = 44

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ (a + l)

⇒ S₄₄ = $\dfrac{44}{2}$ (12 + 98)

= 22 (110)

= 2420.

Hence, S₄₄ = 2420.

Find the sum of first fifteen multiples of 8.

Answer

8, 16, 24,......,(upto 15th term).

The above is an A.P. with common difference equal to 8.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Given,

a = 8

d = 16 - 8 = 8

n = 15

S₁₅ = $\dfrac{15}{2}$ [2(8) + (15 - 1)8]

= $\dfrac{15}{2} \times$ [16 + (14)8]

= $\dfrac{15}{2} \times$ [16 + 112]

= $\dfrac{15}{2} \times$ (128)

= 15 × 64

= 960.

Hence, S₁₅ = 960.

Find the sum of all odd numbers between 100 and 150.

Answer

The odd numbers between 100 and 150 are :

101, 103, 105,........, 149.

Let total number of terms in A.P. be n.

We know that,

∴ a_n = a + (n - 1)d

Given,

a = 101

a_n = 149

d = 103 - 101 = 2

⇒ 149 = 101 + (n - 1)2

⇒ 149 - 101 = (n - 1)2

⇒ 48 = (n - 1)2

⇒ $\dfrac{48}{2}$ = (n - 1)

⇒ 24 = (n - 1)

⇒ n = 24 + 1

⇒ n = 25.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ (a + l)

⇒ S₂₅ = $\dfrac{25}{2}$ (101 + 149)

= $\dfrac{25}{2} \times (250)$

= 25 × 125

= 3125.

Hence, S₂₅ = 3125.

The 4th term of an A.P. is 22 and 15th term is 66. Find the first term and the common difference. Hence, find the sum of the series upto 8 terms.

Answer

Let a be the first term and d be the common difference.

We know that,

∴ a_n = a + (n - 1)d

Given,

The 4th term of an A.P. is 22.

⇒ a₄ = a + (4 - 1)d

⇒ 22 = a + 3d

⇒ a + 3d = 22 .....(1)

Given,

The 15th term of an A.P. is 66.

⇒ a₁₅ = a + (15 - 1)d

⇒ 66 = a + 14d

⇒ a + 14d = 66 .....(2)

Subtracting Equation 1 from Equation 2, we get:

⇒ a + 14d - (a + 3d) = 66 - 22

⇒ a + 14 d - a - 3d = 44

⇒ 11d = 44

⇒ d = $\dfrac{44}{11}$

⇒ d = 4.

Substituting value of d in equation (1), we get :

⇒ a + 3(4) = 22

⇒ a + 12 = 22

⇒ a = 22 - 12

⇒ a = 10.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₈ = $\dfrac{8}{2}$ [2(10) + (8 - 1)4]

= 4[20 + (7)4]

= 4[20 + 28]

= 4 × (48)

= 192.

Hence, a = 10, d = 4, S₈ = 192.

In an Arithmetic Progression (A.P.), the fourth and sixth terms are 8 and 14 respectively. Find the :

(i) first term

(ii) common difference

(iii) sum of the first 20 terms.

Answer

Let a be the first term and d be the common difference.

We know that,

∴ a_n = a + (n - 1)d

Given,

The 4th term of an A.P. is 8.

⇒ a₄ = a + (4 - 1)d

⇒ 8 = a + 3d

⇒ a + 3d = 8 ....(1)

Given,

The 6th term of an A.P. is 14.

⇒ a₆ = a + (6 - 1)d

⇒ 14 = a + 5d

⇒ a + 5d = 14 ....(2)

Subtracting Equation (1) from Equation (2), we get:

⇒ a + 5d - (a + 3d) = 14 - 8

⇒ a + 5d - a - 3d = 6

⇒ 2d = 6

⇒ d = $\dfrac{6}{2}$

⇒ d = 3.

Substituting value of d in equation (1):

⇒ a + 3(3) = 8

⇒ a + 9 = 8

⇒ a = 8 - 9

⇒ a = -1.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Now we have,

a = -1

d = 3

n = 20

⇒ S₂₀ = $\dfrac{20}{2}$ [2(-1) + (20 - 1)3]

= 10 [-2 + (19)3]

= 10 [-2 + 57]

= 10 × (55)

= 550.

Hence, a = -1, d = 3 and S₂₀ = 550.

The sum of the first three terms of an Arithmetic Progression (A.P) is 42 and the product of the first and third term is 52.Find the first term and the common difference.

Answer

Let the three consecutive terms of the A.P. be :

a - d, a, a + d

Given,

The sum of the first three terms is 42.

⇒ a - d + a + a + d = 42

⇒ 3a = 42

⇒ a = $\dfrac{42}{3}$

⇒ a = 14.

Given,

The product of the first term and the third term is 52:

⇒ (a - d)(a + d) = 52

⇒ (14 - d)(14 + d) = 52

⇒ (14)² - d² = 52

⇒ 196 - d² = 52

⇒ 196 - 52 = d²

⇒ 144 = d²

⇒ d² = 144

⇒ d = $\sqrt{144}$

⇒ d = 12 or -12

Case 1: a = 14 and d = 12

First term = a - d = 14 - 12 = 2

Case 2: a = 14 and d = -12

First term = a - d = 14 - (-12) = 26

Hence, first term = 2 and common difference = 12 or first term = 26 and common difference = -12.

If the 6th term of an A.P is equal to four times its first term, and the sum of first six terms is 75, find the first term and the common difference.

Answer

Let a be the first term and d be the common difference.

We know that,

∴ a_n = a + (n - 1)d

Given,

6th term is equal to four times the first term.

⇒ a₆ = 4a₁

⇒ a + (6 - 1)d = 4a

⇒ 5d = 4a - a

⇒ 5d = 3a .........(1)

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Given,

Sum of first six terms is 75.

⇒ S₆ = 75

⇒ $\dfrac{6}{2}$ [2a + (6 - 1)d] = 75

⇒ 3[2a + 5d] = 75

⇒ 2a + 5d = $\dfrac{75}{3}$

⇒ 2a + 5d = 25 ....(2)

Substituting value of 5d from equation (1) in (2), we get:

⇒ 2a + 3a = 25

⇒ 5a = 25

⇒ a = $\dfrac{25}{5}$

⇒ a = 5.

Substitute a = 5 into Equation 1, we get :

⇒ 5d = 3a

⇒ 5d = 3(5)

⇒ 5d = 15

⇒ d = $\dfrac{15}{5}$

⇒ d = 3.

Hence, a = 5 and d = 3.

The 5th term and the 9th term of an Arithmetic Progression are 4 and −12 respectively. Find:

(i) the first term

(ii) common difference

(iii) sum of first 16 terms of the AP.

Answer

Let a be the first term and d be the common difference.

We know that,

∴ a_n = a + (n - 1)d

Given,

The 5th term of an A.P. is 4.

⇒ a + (5 - 1)d = 4

⇒ a + 4d = 4 ....(1)

Given,

The 9th term of an A.P. is -12.

⇒ a + (9 - 1)d = -12

⇒ a + 8d = -12 ....(2)

Subtracting Equation (1) from Equation (2):

⇒ a + 8d - (a + 4d) = -12 - 4

⇒ a + 8d - a - 4d = -16

⇒ 4d = -16

⇒ d = $\dfrac{-16}{4}$

⇒ d = -4.

Substituting d = -4 in Equation (1) :

⇒ a + 4(-4) = 4

⇒ a - 16 = 4

⇒ a = 4 + 16

⇒ a = 20.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Now we have,

a = 20

d = -4

n = 16

⇒ S₁₆ = $\dfrac{16}{2}$ [2(20) + (16 - 1)-4]

= 8[40 + (15)-4]

= 8[40 - 60]

= 8.(-20)

= -160.

Hence, a = 20, d = -4 and S₁₆ = -160.

Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60, … is 300? Hence find the sum of all the terms of the Arithmetic Progression (A.P.).

Answer

The given A.P. is 15, 30, 45, 60,......

a = 15

d = 30 - 15 = 15

a_n = 300

We know that,

∴ a_n = a + (n - 1)d

⇒ 300 = 15 + (n - 1)15

⇒ 300 - 15 = (n - 1) 15

⇒ 285 = (n - 1)15

⇒ $\dfrac{285}{15}$ = n - 1

⇒ 19 = n - 1

⇒ n = 19 + 1

⇒ n = 20

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$(a + l)

Now we have,

a = 15

n = 20

l = 300

⇒ S₂₀ = $\dfrac{20}{2}$ (15 + 300)

= 10 (315)

= 3150

Hence, S₂₀ = 3150.

A sum of ₹ 2,800 is to be used to award four prizes. If each prize after the first is ₹ 200 less than the preceding prize, find the value of each of these prizes.

Answer

Given,

Each prize is ₹ 200 less than the preceding prize.

Prizes : a, a - 200, a - 400, a - 600.

The above is an A.P., wth first term = a and common difference = ₹ -200.

Total prize money = ₹ 2,800.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow 2800 = \dfrac{4}{2} \times [2 \times a + (4 - 1) \times -200] \\[1em] \Rightarrow 2800 = 2 \times [2a - 600] \\[1em] \Rightarrow 2800 = 4a - 1200 \\[1em] \Rightarrow 4a = 2800 + 1200 \\[1em] \Rightarrow 4a = 4000 \\[1em] \Rightarrow a = \dfrac{4000}{4} = ₹ 1,000.$

a - ₹ 200 = ₹ 1,000 - ₹ 200 = ₹ 800

a - ₹ 400 = ₹ 1,000 - ₹ 400 = ₹ 600

a - ₹ 600 = ₹ 1,000 - ₹ 600 = ₹ 400.

Hence, 1st Prize = ₹ 1,000, 2nd Prize = ₹ 800, 3rd Prize = ₹ 600, 4th Prize = ₹ 400.

The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 8000 sets in 6th year, and 11300 in 9th year. Find the production in:

(i) first year

(ii) 8th year

(iii) total production in 6 years.

Answer

(i) Given,

The production of TV sets in a factory increases uniformly by a fixed number every year. So the production is in A.P.

Let the TV sets produced in first year be a and difference in production of each year be d.

We know that,

∴ a_n = a + (n - 1)d

Given,

The sets produced in 6 years is 8000.

⇒ a + (6 - 1)d = 8000

⇒ a + 5d = 8000 ....(1)

The sets produced in the 9th year is 11300.

⇒ a + (9 - 1)d = 11300

⇒ a + 8d = 11300 ....(2)

Subtract Equation (1) from Equation (2), we get:

⇒ a + 8d - (a + 5d) = 11300 - 8000

⇒ 3d = 3300

⇒ d = $\dfrac{3300}{3}$

⇒ d = 1100.

Substitute d = 1100 into Equation 1 :

⇒ a + 5(1100) = 8000

⇒ a + 5500 = 8000

⇒ a + 5500 = 8000

⇒ a = 8000 - 5500

⇒ a = 2500.

Hence, the production in the first year is 2500 sets.

(ii) Solving,

⇒ a₈ = a + 7d

= 2500 + 7(1100)

= 2500 + 7700

= 10200.

Hence, production in 8th year = 10200.

(iii) Total production in 6 years :

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₆ = $\dfrac{6}{2}$ [2(2500) + (6 - 1)1100]

= 3[5000 + (5)1100]

= 3[5000 + 5500]

= 3(10500)

= 31500.

Hence, total production in 6 years = 31500.

200 logs are stacked so that there are 20 logs in the bottom row, 19 logs in the next row, 18 in the next, and so on. How many rows are formed and how many logs are there in the top row?

Answer

The number of logs in each row forms an Arithmetic Progression (A.P.):

20, 19, 18,......

Total sum of logs = 200.

S_n = 200

Common difference (d) = 19 - 20 = -1

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ 200 = $\dfrac{n}{2}$ [2(20) + (n - 1)(-1)]

⇒ 200 × 2 = n[40 - n + 1]

⇒ 400 = n(41 - n)

⇒ 400 = 41n - n²

⇒ n² - 41n + 400 = 0

⇒ n² - 16n - 25n + 400 = 0

⇒ n(n - 16) - 25(n - 16) = 0

⇒ (n - 25)(n - 16) = 0

⇒ (n - 25) = 0 or (n - 16) = 0

⇒ n = 16 or n = 25.

Case 1 :

If n = 25,

logs in the 25th row a₂₅:

⇒ a₂₅ = a + (25 - 1)d

= 20 + (24)(-1)

= 20 - 24

= -4 (the number of logs cannot be negative).

Thus, n ≠ 25.

If n = 16

logs in the 16th row a₁₆:

⇒ a₁₆ = a + (16 - 1)d

= 20 + (15)(-1)

= 20 - 15

= 5.

Hence, no. of rows = 16 and the top row has 5 logs.

In a flower bed there are 43 rose plants in the first row, 41 in the second, 39 in the third and so on. There are 11 rose plants in the last row. How many rows are there and how many rose plants are there in the bed?

Answer

Given,

The number of rose plants in each row forms an Arithmetic Progression (A.P.):

43, 41, 39,......., 11.

a = 43

l = 11

d = 43 - 41 = -2

We know that,

∴ a_n = a + (n - 1)d

⇒ 11 = 43 + (n - 1)(-2)

⇒ 11 - 43 = (n - 1)(-2)

⇒ -32 = (n - 1)(-2)

⇒ $\dfrac{-32}{-2}$ = (n - 1)

⇒ n - 1 = 16

⇒ n = 16 + 1

⇒ n = 17

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$(a + l)

Total number of rose plants:

⇒ S₁₇ = $\dfrac{17}{2}$(43 + 11)

= $\dfrac{17}{2}$(43 + 11)

= 15 × $\dfrac{54}{2}$

= 17 × 27

= 459.

Hence, number of rows = 17, total number of rose plants in the bed is 459.

A man saved ₹ 33,000 in 10 months. In each month after the first, he saves ₹ 100 more than he did in the preceding month. How much did he save in the first month?

Answer

Let the man saved ₹ a in first month.

The amount the man saved each month forms an Arithmetic Progression (A.P.) where :

Total sum saved S_n = ₹ 33,000.

Number of months (n) = 10

He saves ₹ 100 more than he did in the preceding month.

Thus, d = 100

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$[2a + (n - 1)d]

⇒ 33000 = $\dfrac{10}{2}$ [2a + (10 - 1)100]

⇒ 33000 = 5[2a + (9)100]

⇒ $\dfrac{33000}{5}$ = [2a + 900]

⇒ 6600 = [2a + 900]

⇒ 6600 - 900 = 2a

⇒ 2a = 5700

⇒ a = $\dfrac{5700}{2}$

⇒ a = ₹ 2,850.

Hence, the amount man saved in first month is ₹ 2,850.

The general term of an A.P. whose first term is a and common difference is d, is given by:

1. T_n = 2a + (n − 1)d

2. T_n = $\dfrac{n}{2}[2a + (n − 1)d]$

3. T_n = a + (n − 1)d

4. T_n = $\dfrac{n}{2}[a + (n − 1)d]$

Answer

n^th term of an A.P. is given by,

T_n = a + (n - 1)d

Hence, option 3 is the correct option.

The nth term from the end of an A.P., whose first term, last term and common difference are a, l and d respectively, is given by:

1. l + (n + 1)d

2. l − (n − 1)d

3. l + (n − 1)d

4. l − (n + 1)d

Answer

Since, we need to calculate term from end.

So, first term will be equal to last term and common difference will be negative of the original difference.

∴ T_end = l + (n - 1)(-d)

= l - (n - 1)d.

Hence, option 2 is the correct option.

Sum of n terms of an A.P. whose first term is a and common difference is d, is given by:

1. S_n = $\dfrac{n}{2}[a + (n − 1)d]$

2. S_n = $\dfrac{n}{2}[2a + (n + 1)d]$

3. S_n = $\dfrac{n}{2}[2a − (n − 1)d]$

4. S_n = $\dfrac{n}{2}[2a + (n − 1)d]$

Answer

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$[2a + (n - 1)d]

Hence, option 4 is the correct option.

Sum of n terms of an A.P. whose first term and last term are a and l respectively is given by:

1. S_n = $\dfrac{n}{2}(a + l)$

2. S_n = $\dfrac{n}{2}(a − l)$

3. S_n = $\dfrac{n}{2}(2a + l)$

4. S_n = $\dfrac{n}{2}[a + (n − 1)l]$

Answer

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$(a + l)

Hence, option 1 is the correct option.

The sum of first n natural numbers is:

1. $\dfrac{n(n − 1)}{2}$

2. $\dfrac{n(n + 1)}{2}$

3. $\dfrac{n(n + 2)}{2}$

4. $\dfrac{n(n − 2)}{2}$

Answer

The sequence of natural numbers 1, 2, 3,....., n.

a = 1

l = n

no. of terms = n

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ (a + l)

= $\dfrac{n}{2}$ (1 + n)

= $\dfrac{n(n + 1)}{2}$

Hence, option 2 is the correct option.

The sum of first 50 natural numbers is:

1. 1050

2. 1175

3. 1225

4. 1275

Answer

Sequence : 1, 2, 3, ......, 50.

First term (a) = 1

Common difference (d) = 2 - 1 = 1

By formula,

Sum of n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

$= \dfrac{50}{2} \times [2 \times 1 + (50 - 1) \times 1] \\[1em] = 25 \times [2 + 49] \\[1em] = 25 \times 51 \\[1em] = 1275.$

Hence, option 4 is the correct option.

The nth term of an Arithmetic Progression (A.P.) is 2n + 5. The 10th term is :

1. 7

2. 15

3. 25

4. 45

Answer

Given,

nth term of A.P. :

∴ T_n = 2n + 5

⇒ T₁₀ = 2(10) + 5

= 20 + 5

= 25.

Hence, option 3 is the correct option.

The first term of an A.P. is 7 and the common difference is 3. The general term of the A.P. is:

1. T_n = 3n − 4

2. T_n = 2n + 5

3. T_n = 3n + 4

4. None of these

Answer

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

Given,

a = 7, d = 3

T_n = 7 + (n - 1)3

= 7 + 3n - 3

= 3n + 4.

Hence, option 3 is the correct option.

The 24th term of the A.P. −1, 3, 7, 11, … is:

1. 83

2. 87

3. 91

4. 95

Answer

The given Arithmetic Progression (A.P.) is

-1, 3, 7, 11,.....

a = -1

d = 3 - (-1) = 4

n = 24

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₂₄ = -1 + (24 - 1)(4)

= -1 + (23)4

= -1 + 92

= 91.

Hence, option 3 is the correct option.

If 70, 75, 80, 85 are the first four terms of an arithmetic progression, then the 10th term is:

1. 35

2. 25

3. 115

4. 105

Answer

The given Arithmetic Progression (A.P.) is

70, 75, 80, 85, ....

a = 70

d = 75 - 70 = 5

n = 10

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₁₀ = 70 + (10 - 1)5

= 70 + (9)5

= 70 + 45

= 115.

Hence, option 3 is the correct option.

The A.P. 6, 13, 20, …, 216 has 31 terms. The middle term of the A.P. is :

1. 91

2. 97

3. 107

4. 111

Answer

Given,

A.P. 6, 13, 20, …, 216

n = 31

Middle term of A.P. = $\dfrac{n + 1}{2}$

= $\dfrac{31 + 1}{2}$

= $\dfrac{32}{2}$

= 16th term.

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

Now we have,

a = 6

d = 13 - 6 = 7

n = 16

⇒ T₁₆ = 6 + (16 - 1)(7)

= 6 + (15)7

= 6 + 105

= 111.

Hence, option 4 is the correct option.

Which term of the A.P. 7, 13, 19, 25,..... is 241?

1. 40th

2. 36th

3. 44th

4. 45th

Answer

Given,

A.P. : 7, 13, 19, 25,.......

a = 7

d = 13 - 7 = 6

Let nth term be 241.

T_n = 241

⇒ a + (n - 1)d = 241

⇒ 241 = 7 + (n - 1)6

⇒ 241 - 7 = (n - 1)6

⇒ 234 = (n - 1)6

⇒ $\dfrac{234}{6}$ = n - 1

⇒ n - 1 = 39

⇒ n = 39 + 1

⇒ n = 40.

Hence, option 1 is the correct option.

Which term of the A.P. 11, 8, 5, 2, ..... is −148 ?

1. 52nd

2. 54th

3. 55th

4. 57th

Answer

Given,

A.P. : 11, 8, 5, 2, .......

a = 11

d = 8 - 11 = -3

Let nth term of the A.P. be -148.

T_n = -148

⇒ a + (n - 1)d = -148

⇒ -148 = 11 + (n - 1)(-3)

⇒ -148 - 11 = (n - 1)(-3)

⇒ -159 = (n - 1)(-3)

⇒ $\dfrac{-159}{-3}$ = n - 1

⇒ n - 1 = 53

⇒ n = 53 + 1

⇒ n = 54.

Hence, option 2 is the correct option.

How many three-digit numbers are divisible by 7?

1. 128

2. 124

3. 136

4. 132

Answer

The three-digit numbers divisible by 7 form an Arithmetic Progression (A.P.) :

105, 112, 119, ......,994.

a = 105

l = 994

d = 7

Let no. of terms be n.

∴ T_n = a + (n - 1)d

⇒ 994 = 105 + (n - 1)7

⇒ 994 - 105 = (n - 1)7

⇒ 889 = (n - 1)7

⇒ $\dfrac{889}{7}$ = n - 1

⇒ 127 = n - 1

⇒ n = 127 + 1

⇒ n = 128.

Hence, option 1 is the correct option.

How many numbers lying between 20 and 200 are divisible by 4?

1. 43

2. 44

3. 45

4. 46

Answer

The numbers divisible by 4 lying between 20 and 200 form an Arithmetic Progression (A.P.).

24, 28, ....., 196.

a = 24

l = 196

d = 4

Let no. of terms be n.

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ 196 = 24 + (n - 1)4

⇒ 196 - 24 = (n - 1)4

⇒ 172 = (n - 1)4

⇒ $\dfrac{172}{4}$ = n - 1

⇒ 43 = n - 1

⇒ n = 43 + 1

⇒ n = 44.

Hence, option 2 is the correct option.

The common difference of the A.P. $\dfrac{1}{k}, \dfrac{1 - k}{k}, \dfrac{1 - 2k}{k}$, ...... is :

1. −k

2. k

3. 1

4. −1

Answer

Given,

A.P. : $\dfrac{1}{k}, \dfrac{1 - k}{k}, \dfrac{1 - 2k}{k},$ .........

a = $\dfrac{1}{k}$

$\Rightarrow d = \dfrac{1 - k}{k} - \dfrac{1}{k} \\[1em] \Rightarrow d = \dfrac{1 - k - 1}{k} \\[1em] \Rightarrow d = \dfrac{-k}{k} \\[1em] \Rightarrow d = -1.$

Hence, option 4 is the correct option.

If the nth term of an A.P. is given by (3n + 2), then the sum of its first three terms is :

1. 21

2. 24

3. 27

4. 32

Answer

The nth term of the A.P. is given by,

T_n = 3n + 2

T₁ = 3(1) + 2 = 5

T₂ = 3(2) + 2 = 8

T₃ = 3(3) + 2 = 11

Sum = 5 + 8 + 11 = 24.

Hence, option 2 is the correct option.

The last term of the A.P., 5, 12, 19, ..... having 60 terms is :

1. 406

2. 412

3. 416

4. 418

Answer

Given,

A.P. : 5, 12, 19, .......

a = 5

d = 12 - 5 = 7

n = 60

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₆₀ = 5 + (60 - 1)7

= 5 + (59) × 7

= 5 + 413

= 418.

Hence, option 4 is the correct option.

The common difference of the A.P. : $\dfrac{1}{3}, \dfrac{(1 − 3p)}{3}, \dfrac{(1 − 6p)}{3}$, ... is:

1. p

2. −p

3. $\dfrac{1}{3}$

4. $\dfrac{-1}{3}$

Answer

In the above A.P.,

$\Rightarrow d = \dfrac{(1 − 3p)}{3} - \dfrac{1}{3} \\[1em] \Rightarrow d = \dfrac{(1 − 3p) - 1}{3} \\[1em] \Rightarrow d = \dfrac{-3p}{3} \\[1em] \Rightarrow d = -p.$

Hence, option 2 is the correct option.

The next term of the A.P. $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}$, ... is:

1. $\sqrt{72}$

2. $\sqrt{68}$

3. $\sqrt{75}$

4. $\sqrt{56}$

Answer

In the above A.P. next terms is the 5th term.

a = $\sqrt{3}$

n = 5

$\Rightarrow d = \sqrt{12} - \sqrt{3} \\[1em] \Rightarrow d = \sqrt{4 \times 3} - \sqrt3 \\[1em] \Rightarrow d = 2\sqrt{3} - \sqrt3 \\[1em] \Rightarrow d = \sqrt3.$

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

$\Rightarrow T_5 = \sqrt3 + (5 - 1)\sqrt3 \\[1em] = \sqrt3 + 4\sqrt3 \\[1em] = 5\sqrt3 \\[1em] = \sqrt{5^2 \times 3} \\[1em] = \sqrt{75}.$

Hence, option 3 is the correct option.

The first term of an A.P. is p and its common difference is q. The 10th term of the A.P. is :

1. p − 9q

2. p + 10q

3. p − 10q

4. p + 9q

Answer

Given,

First term = p

Common difference = q

n = 10

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

⇒ T₁₀ = p + (10 - 1)q

= p + 9q.

Hence, option 4 is the correct option.

The nth term of the A.P. $\dfrac {1}{m}, \dfrac{(1 + m)}{m}, \dfrac{(1 + 2m)}{m}$, ... is:

1. $\dfrac{m(n − 1) + 1}{m}$

2. $\dfrac{m(n + 1) − 1}{m}$

3. $\dfrac{m(n − 1) − 1}{m}$

4. None of these

Answer

Given,

a = $\dfrac{1}{m}$

$\Rightarrow d = \dfrac{(1 + m)}{m} - \dfrac {1}{m} \\[1em] \Rightarrow d = \dfrac{(1 + m) - 1}{m} \\[1em] \Rightarrow d = \dfrac{m}{m} \\[1em] \Rightarrow d = 1.$

We know that,

nth term of A.P. :

∴ T_n = a + (n - 1)d

$\Rightarrow T_n = \dfrac{1}{m} + (n - 1) \times 1 \\[1em] = \dfrac{1 + m(n - 1)}{m} \\[1em] = \dfrac{m(n - 1) + 1}{m}.$

Hence, option 1 is the correct option.

For what value of p are 2p + 1, 13, 5p − 3 three consecutive terms of an A.P. ?

1. 0

2. 1

3. 2

4. 4

Answer

Given,

2p + 1, 13, 5p − 3

In an A.P., the middle term is the average of the first and third terms.

$\Rightarrow 13 = \dfrac{(2p + 1) + (5p - 3)}{2} \\[1em] \Rightarrow 13 = \dfrac{7p - 2}{2} \\[1em] \Rightarrow 13 \times 2 = 7p - 2 \\[1em] \Rightarrow 26 = 7p - 2 \\[1em] \Rightarrow 26 + 2 = 7p \\[1em] \Rightarrow 28 = 7p \\[1em] \Rightarrow 7p = 28 \\[1em] \Rightarrow p = \dfrac{28}{7} \\[1em] \Rightarrow p = 4.$

Hence, option 4 is the correct option.

For what value of k will 2k + 1, 3k + 3, and 5k − 1 be three consecutive terms of an A.P.?

1. 1

2. 2

3. 4

4. 6

Answer

We are given three consecutive terms of an A.P.:

2k + 1, 3k + 3, 5k − 1

In an arithmetic progression, the difference between consecutive terms is the same.

⇒ (3k + 3) - (2k + 1) = (5k - 1) - (3k + 3)

⇒ 3k + 3 - 2k - 1 = 5k - 1 - 3k - 3

⇒ k + 2 = 2k - 4

⇒ 2 + 4 = 2k - k

⇒ k = 6.

Hence, option 4 is the correct option.

If the sum of first n terms of an A.P. is S_n = 5n² + 3n, then its common difference is:

1. 8

2. 10

3. 18

4. 26

Answer

Given,

S_n = 5n² + 3n

We know that,

nth term of A.P. :

∴ T_n = S_n - S_{n - 1}

= 5n² + 3n - [5(n - 1)² + 3(n - 1)]

= 5n² + 3n - [5(n² - 2n + 1) + 3n - 3]

= 5n² + 3n - [5n² - 10n + 5 + 3n - 3]

= 5n² + 3n - 5n² + 10n - 5 - 3n + 3

= 10n - 2

d = a_n - a_{n - 1}

= 10n - 2 - [10(n - 1) - 2]

= 10n - 2 - [10n - 10 - 2]

= 10n - 2 - 10n + 10 + 2

= 10.

Hence, option 2 is the correct option.

The first and the last terms of an A.P. are 1 and 11 respectively. If the sum of its terms is 36, then the number of terms is :

1. 6

2. 7

3. 8

4. 9

Answer

Given,

a = 1

l = 11

Let no. of terms be n.

S_n = 36

We know that,

⇒ S_n = $\dfrac{n}{2}$ (a + l)

⇒ 36 = $\dfrac{n}{2}$ (1 + 11)

⇒ 36 × 2 = n(12)

⇒ n = $\dfrac{36 \times 2}{12}$

⇒ n = 6.

Hence, option 1 is the correct option.

The sum of first 40 positive integers divisible by 6 is :

1. 2460

2. 3640

3. 4920

4. 4860

Answer

Sequence :

6, 12, 18, ......., upto 40th term.

The above sequence is an A.P. with,

a = 6

d = 6

n = 40

We know that,

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₄₀ = $\dfrac{40}{2}$ [2(6) + (40 - 1)6]

= 20[12 + (39)6]

= 20(12 + 234)

= 20 × (246)

= 4920.

Hence, option 3 is the correct option.

The sum of first 30 odd natural numbers is:

1. 800

2. 900

3. 729

4. 1249

Answer

1, 3, 5, 7, ......, 30th term.

The odd natural numbers form an Arithmetic Progression (A.P.) :

a = 1

d = 3 - 1 = 2

n = 30

We know that,

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₃₀ = $\dfrac{30}{2}$ [2(1) + (30 - 1)2]

= 15[2 + (29)2]

= 15[2 + 58]

= 15 × (60)

= 900.

Hence, option 2 is the correct option.

The 7th term of an A.P. is 4 and its common difference is −4. The first term of the A.P. is :

1. 32

2. 28

3. 24

4. 36

Answer

Given,

a₇ = 4

d = -4

n = 7

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₇ = a + (7 - 1)(-4)

⇒ 4 = a + (6)(-4)

⇒ 4 = a - 24

⇒ 24 + 4 = a

⇒ a = 28.

Hence, option 2 is the correct option.

The sum of first n terms of an A.P. is (4n² + 2n). The nth term of the A.P. is :

1. (6n − 2)

2. (8n − 2)

3. (6n + 2)

4. (8n + 2)

Answer

We know that,

S_n = 4n² + 2n

T_n = S_n - S_{n - 1}

= 4n² + 2n - [4(n - 1)² + 2(n - 1)]

= 4n² + 2n - [4(n² - 2n + 1) + 2n - 2]

= 4n² + 2n - [4n² - 8n + 4 + 2n - 2]

= 4n² + 2n -[4n² - 6n + 2]

= 4n² + 2n - 4n² + 6n - 2

= 8n - 2.

Hence, option 2 is the correct option.

The first term of an A.P. is 7 and its 13th term is 35. The common difference of the A.P. is :

1. $\dfrac{5}{3}$

2. $\dfrac{7}{2}$

3. $\dfrac{8}{3}$

4. $\dfrac{7}{3}$

Answer

Given,

a = 7

⇒ a₁₃ = 35

⇒ a + (13 - 1)d = 35

⇒ 7 + 12d = 35

⇒ 12d = 35 - 7

⇒ 12d = 28

⇒ d = $\dfrac{28}{12}$

⇒ d = $\dfrac{7}{3}$.

Hence, option 4 is the correct option.

There are total 9 terms in an A.P. If the last term and the sum of all the terms are 28 and 144 respectively, then the first term is :

1. 1

2. 7

3. 4

4. 5

Answer

We know that,

S_n = $\dfrac{n}{2}$(a + l)

Given,

n = 9

l = 28

S_n = 144

⇒ S₉ = $\dfrac{9}{2}$(a + 28)

⇒ 144 = $\dfrac{9}{2}$(a + 28)

⇒ $\dfrac{144 \times 2}{9}$ = (a + 28)

⇒ 32 = (a + 28)

⇒ 32 - 28 = a

⇒ a = 4.

Hence, option 3 is the correct option.

The 5th term of an A.P. is −3 and its common difference is −4. The sum of its first 10 terms is :

1. −50

2. −40

3. −60

4. −30

Answer

Given,

a₅ = -3

d = -4

n = 10

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₅ = a + (5 - 1)(-4)

⇒ -3 = a + (4)(-4)

⇒ -3 = a - 16

⇒ a = 16 - 3

⇒ a = 13.

We know that,

⇒ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₁₀ = $\dfrac{10}{2}$ [2(13) + (10 - 1)(-4)]

= 5[26 + (9)(-4)]

= 5[26 - 36]

= 5(-10)

= -50.

Hence, option 1 is the correct option.

The 7th term of an A.P. is −1 and its 16th term is 17. The nth term of the A.P. is :

1. (3n + 12)

2. (2n − 5)

3. (3n + 5)

4. (2n − 15)

Answer

We know that,

a_n = a + (n - 1)d

The 7th term of an A.P. is −1.

a + 6d = -1 ......(1)

The 16th term of an A.P. is 17.

a + 15d = 17 ......(2)

Subtract Equation (1) from Equation (2) :

⇒ a + 15d - (a + 6d) = 17 - (-1)

⇒ a + 15d - a - 6d = 18

⇒ 9d = 18

⇒ d = $\dfrac{18}{9}$

⇒ d = 2.

Substituting d = 2 into Equation (1), we get :

⇒ a + 6(2) = -1

⇒ a + 12 = -1

⇒ a = -1 - 12

⇒ a = -13.

Now,

⇒ a_n = a + (n - 1)d

⇒ a_n = -13 + (n - 1)2

= -13 + 2n - 2

= 2n - 15.

Hence, option 4 is the correct option.

If the sum of first p terms of an A.P. is ap² + bp, then its common difference is :

1. a

2. 3a + b

3. a + b

4. 2a

Answer

Given,

⇒ S_p = ap² + bp

⇒ T_p = S_p - S_{p - 1}

= ap² + bp - [a(p - 1)² + b(p - 1)]

= ap² + bp - [a(p² - 2p + 1) + bp - b]

= ap² + bp - [ap² - 2ap + a + bp - b]

= ap² + bp - ap² + 2ap - a - bp + b

= 2ap - a + b.

d = T_p - T_{p - 1}

= 2ap - a + b - [2a(p - 1) - a + b]

= 2ap - a + b - [2ap - 2a - a + b]

= 2ap - a + b - 2ap + 2a + a - b

= 2a.

Hence, option 4 is the correct option.

The first term of an A.P. is a and nth term is b, then the common difference of the A.P. is :

1. $\dfrac{b − a}{n}$

2. $\dfrac{b − a}{n − 1}$

3. $\dfrac{b + a}{n − 1}$

4. $\dfrac{b − a}{n + 1}$

Answer

Given,

First term = a

nth term = b

We know that,

a_n = a + (n - 1)d

⇒ b = a + (n - 1)d

⇒ b - a = (n - 1)d

⇒ d = $\dfrac{b - a}{n - 1}$.

Hence, option 2 is the correct option.

The first three terms of an A.P. are 3y − 1, 3y + 5 and 5y + 1 respectively. Then, the value of y is :

1. 1

2. 5

3. 8

4. 3

Answer

Given,

The first three terms of an A.P. are 3y − 1, 3y + 5 and 5y + 1.

The difference between consecutive terms must be equal.

⇒ 3y + 5 - (3y - 1) = 5y + 1 - (3y + 5)

⇒ 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5

⇒ 6 = 2y - 4

⇒ 2y = 6 + 4

⇒ 2y = 10

⇒ y = $\dfrac{10}{2}$

⇒ y = 5.

Hence, option 2 is the correct option.

The 8th term from the end of the A.P. 7, 10, 13, ..., 184 is:

1. 157

2. 160

3. 163

4. 166

Answer

Given,

a = 7

d = 10 - 7 = 3

a_n = 184

We know that,

⇒ a_n = a + (n - 1)d

⇒ a_n = 7 + (n - 1)(3)

⇒ 184 = 7 + (n - 1)(3)

⇒ 184 - 7 = (n - 1)(3)

⇒ 177 = (n - 1)(3)

⇒ $\dfrac{177}{3}$ = (n - 1)

⇒ n - 1 = 59

⇒ n = 59 + 1

⇒ n = 60.

The 8th term from the end is the (n - 8 + 1)th term from the beginning.

= 60 - 8 + 1

= 53.

⇒ a₅₃ = 7 + (53 - 1)3

= 7 + (52)(3)

= 7 + 156

= 163.

Hence, option 3 is the correct option.

If a = 3, n = 8 and S_n = 192, then the common difference of the A.P. is:

1. 4

2. 5

3. 6

4. 7

Answer

a = 3

n = 8

S_n = 192

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ 192 = $\dfrac{8}{2}$ [2(3) + (8 - 1)d]

⇒ 192 = 4[6 + 7d]

⇒ $\dfrac{192}{4}$ = [6 + 7d]

⇒ 48 = [6 + 7d]

⇒ 48 - 6 = 7d

⇒ 42 = 7d

⇒ d = $\dfrac{42}{7}$

⇒ d = 6.

Hence, option 3 is the correct option.

Case Study I

A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A, of radii 0.5 cm, 1 cm, 1.5 cm, 2 cm, ...... as shown in the given figure.

Based on this information, answer the following questions:

(Take π = $\dfrac{22}{7}$)

1. What is the radius of the 9th semicircle?
   (a) 4 cm
   (b) 4.5 cm
   (c) 3.5 cm
   (d) 5 cm

2. The length of the 15th semicircle is :
   (a) 8π cm
   (b) 7π cm
   (c) 7.5π cm
   (d) 6.5π cm

3. The difference of lengths of the 19th and 12th semicircles is :
   (a) 3.5π cm
   (b) 3 cm
   (c) 4π cm
   (d) 4.5π cm

4. The total length of the spiral made up of first 7 consecutive semicircles is:
   (a) 38 cm
   (b) 42 cm
   (c) 44 cm
   (d) 46 cm

5. The lengths of the semicircles form an A.P. What is the common difference of this A.P.?
   (a) 0.5π cm
   (b) π cm
   (c) 0.25π cm
   (d) 1.25π cm

Answer

1. 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ....

The radius are in A.P. with

a = 0.5 cm

d = 1.0 - 0.5 = 0.5 cm

n = 9

We know that,

⇒ a_n = a + (n - 1)d

⇒ r₉ = 0.5 + (9 - 1)0.5

= 0.5 + (8)0.5

= 0.5 + 4

= 4.5 cm

Hence, option (b) is the correct option.

2. Calculating the radius of 15th semicircle.

We know that,

⇒ a_n = a + (n - 1)d

⇒ r₁₅ = 0.5 + (15 - 1)0.5

= 0.5 + 14(0.5)

= 0.5 + 7

= 7.5 cm

By formula,

Length of a semicircle = πr

= 7.5π cm

Hence, option (c) is the correct option.

3. We know that,

Length of a semicircle = πr

The lengths are: 0.5π cm, 1.0π cm, 1.5π cm, 2.0π cm,.... in an A.P. with,

a = 0.5π cm

d = 1.0π cm - 0.5π cm = 0.5π

The difference of lengths of the 19th and 12th semicircles is:

L₁₉ - L₁₂ = πr₁₉ - πr₁₂

= π(r₁₉ - r₁₂)

We know that,

a_n = a + (n - 1)d

⇒ r₁₉ = 0.5π + (19 - 1)0.5π

= 0.5π + 18(0.5π)

= 0.5π + 9π

= 9.5π cm.

⇒ r₁₂ = 0.5π + (12 - 1)0.5π

= 0.5π + 11(0.5π)

= 0.5π + 5.5π

= 6π cm.

L₁₉ - L₁₂ = (9.5π - 6π)

= 3.5π cm.

Hence, option (a) is the correct option.

4. The lengths are: 0.5π cm, 1.0π cm, 1.5π cm, 2.0π cm,....in an A.P.

a = 0.5π cm

d = 1.0π cm - 0.5π cm = 0.5π

The total length is the sum of the first 7 lengths S₇.

We know that,

⇒ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₇ = $\dfrac{7}{2}$ [2(0.5π) + (7 - 1)(0.5π)]

= 3.5[1π + 6(0.5π)]

= 3.5[1π + 3π]

= 3.5(4π)

= 14π cm.

Total length = 14π cm

= $14 \times \dfrac{22}{7}$

= 44 cm.

Hence, option (c) is the correct option.

5. 0.5π cm, 1.0π cm, 1.5π cm, 2.0π cm, ....

d = 1.0π + 0.5π = 0.5 π cm.

Common difference of the A.P. of lengths = 0.5π cm.

Hence, option (a) is the correct option.

Case Study II

The production of TV sets in a factory increases uniformly by a fixed number every year.
It produced 16000 sets in the 6th year and 22600 in the 9th year.

Based on this information, answer the following questions:

1. The production of the TV sets during the first year was :
   (a) 4000
   (b) 4500
   (c) 5000
   (d) 5500

2. What was the uniform increase in the production of TV sets every year?
   (a) 1800
   (b) 2400
   (c) 1600
   (d) 2200

3. The production of the TV sets during the 8th year was:
   (a) 20000
   (b) 20400
   (c) 21200
   (d) 22800

4. The total production of the TV sets during first 6 years was:
   (a) 56000
   (b) 72000
   (c) 66000
   (d) 63000

5. The average production of the TV sets during first 6 years was:
   (a) 10500
   (b) 11000
   (c) 11500
   (d) 12000

Answer

1. Since, production of TV sets in a factory increases uniformly by a fixed number every year. The production in each year forms an A.P.

Let production of the TV sets during the first year be a and the common difference in production each year be d.

We know that,

a_n = a + (n - 1)d

Given,

Production in the 6th year = 16000.

⇒ a₆ = a + (6 - 1)d

⇒ 16000 = a + 5d

⇒ a + 5d = 16000 ....(1)

Given,

Production in the 9th year = 22600.

⇒ a₉ = a + (9 - 1)d

⇒ 22600 = a + 8d

⇒ a + 8d = 22600 ....(2)

Subtracting Eq. (1) from Eq. (2), we get :

⇒ a + 8d - (a + 5d) = 22600 - 16000

⇒ a + 8d - a - 5d = 6600

⇒ 3d = 6600

⇒ d = $\dfrac{6600}{3}$

⇒ d = 2200.

Substituting d in equation (1), we get :

⇒ a + 5(2200) = 16000

⇒ a + 11000 = 16000

⇒ a = 16000 - 11000

⇒ a = 5000.

Hence, option (c) is the correct option.

2. From question (1),

The uniform increase is the common difference, d.

d = 2200.

Hence, option (d) is the correct option.

3. Given,

The production of TV sets in each year forms an A.P.: 5000, 7200, 9400, .....

a = 5000

n = 8

d = 2200

We know that,

a_n = a + (n - 1)d

⇒ a₈ = 5000 + (8 - 1)2200

= 5000 + 7(2200)

= 5000 + 15400

= 20400.

Production of the TV sets during the 8th year = 20400.

Hence, option (b) is the correct option.

4. Given,

The production of TV sets in each year forms an A.P.: 5000, 7200, 9400.....

a = 5000

l = 16000

n = 6

We know that,

S_n = $\dfrac{n}{2}$ (a + l)

⇒ S₆ = $\dfrac{6}{2}$ (5000 + 16000)

= 3 × (21000)

= 63000.

Total production of the TV sets during first 6 years = 63000

Hence, option (d) is the correct option.

5. From part 4 we have,

S₆ = 63000

The average production is the total production divided by the number of years:

Average = $\dfrac{S_6}{6}$

= $\dfrac{63000}{6}$

= 10500.

Hence, option (a) is the correct option.

Case Study III

200 logs are stacked in the following manner:
20 logs in the bottom row, 19 in the next row, 18 in the next row and so on.

Based on this information, answer the following questions:

1. In how many rows these 200 logs are placed?
   (a) 25
   (b) 20
   (c) 16
   (d) 14

2. The number of logs in the top row is:
   (a) 1
   (b) 5
   (c) 3
   (d) 8

3. The number of logs in the 8th row from the bottom is:
   (a) 14
   (b) 11
   (c) 12
   (d) 13

4. Total number of logs in the first six rows from the bottom is:
   (a) 105
   (b) 95
   (c) 85
   (d) 75

5. The number of logs in the 5th row from the top is:
   (a) 8
   (b) 9
   (c) 7
   (d) 10

Answer

1. Given,

Logs stacked in rows form an A.P.: 20, 19, 18.....

S_n = 200

a = 20

d = 19 - 20 = -1

We know that,

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Let 200 logs be stacked in n rows.

⇒ 200 = $\dfrac{n}{2}$ [2(20) + (n - 1)(-1)]

⇒ 200 × 2 = n[40 - n + 1]

⇒ 400 = n[41 - n]

⇒ 400 = 41n - n²

⇒ n² - 41n + 400 = 0

⇒ n² - 16n - 25n + 400 = 0

⇒ n(n - 16) - 25(n - 16) = 0

⇒ (n - 25)(n - 16) = 0

⇒ (n - 25) = 0 or (n - 16) = 0

⇒ n = 25 or n = 16.

If n = 25, a₂₅ = a + 24d = 20 + 24(-1) = -4 (Impossible, the 25th row would have –4 logs).

If n = 16, a₁₆ = a + 15d = 20 + 15(-1) = 5 (Valid).

The number of rows is 16.

Hence, option (c) is the correct option.

2. Given,

Logs stacked in rows form an A.P.: 20, 19, 18, .....

a = 20

n = 16 (16 total rows of logs in the stack i.e top row is 16th.)

d = -1

We know that,

a_n = a + (n - 1)d

⇒ a₁₆ = 20 + (16 - 1)(-1)

= 20 + 15(-1)

= 20 - 15

= 5.

The number of logs in the top row is 5.

Hence, option (b) is the correct option.

3. Given,

Logs stacked in rows form an A.P.: 20, 19, 18.....

a = 20

Total rows = 16

n = 8, (A.P. starts from bottom only)

d = -1

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₈ = 20 + (8 - 1)(-1)

= 20 + 7(-1)

= 20 - 7

= 13.

The number of logs in the 8th row from the bottom is 13.

Hence, option (d) is the correct option.

4. Given,

Logs stacked in rows form an A.P.: 20, 19, 18, .....

a = 20

d = -1

n = 6

We know that,

S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

⇒ S₆ = $\dfrac{6}{2}$ [2(20) + (6 - 1)(-1)]

= 3(40 - 5)

= 3 × (35)

= 105.

Total number of logs in the first six rows from the bottom is 105.

Hence, option (a) is the correct option.

5. The 5th row from the top is the (16 - 5 + 1) = 12th row from the bottom (a₁₂):

Logs stacked in rows form an A.P.: 20, 19, 18.....

We know that,

a_n = a + (n - 1)d

⇒ a₁₂ = 20 + (12 - 1)(-1)

= 20 - 11

= 9.

Hence, option (b) is the correct option.

Assertion (A): The sum of first n terms of the A.P. −1, 5, 11, ... is 3n² − 4n.

Reason (R): The sum of first n terms of an A.P. is given by S_n = $\dfrac{n}{2}$ [2a + (n − 1)d].

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true.

4. Both A and R are false.

Answer

A.P. : -1, 5, 11, ......

Given,

a = -1

d = 5 - (-1) = 6

We know that,

S_n = $\dfrac{n}{2}$ [2a + (n − 1)d]

⇒ S_n = $\dfrac{n}{2}$ [2(-1) + (n − 1)6]

= $\dfrac{n}{2}$ [-2 + (6n − 6)]

= $\dfrac{n}{2}$ (6n − 8)

= $\dfrac{n}{2} \times$ 2(3n − 4)

= n(3n - 4)

= 3n² - 4n.

∴ Assertion (A) is true.

The standard and correct formula for the sum of the first n terms of an A.P.

S_n = $\dfrac{n}{2}$[2a + (n − 1)d]

∴ Reason (R) is true.

Both A and R are true.

Hence, option 3 is the correct option.

Assertion (A): The 10th term from the end of the A.P. 17, 14, 11, ... −40 is −11.

Reason (R): The nth term of an A.P. is given by t_n = a + (n − 1)d.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

Given,

a = 17

a_n = -40

d = 14 - 17 = -3

We know that,

⇒ a_n = a + (n - 1)d

⇒ -40 = 17 + (n - 1)(-3)

⇒ -40 - 17 = (n - 1)(-3)

⇒ -57 = (n - 1)(-3)

⇒ $\dfrac{-57}{-3}$ = (n - 1)

⇒ 19 = n - 1

⇒ n = 19 + 1

⇒ n = 20.

The A.P. has 20 terms.

The 10th term from the end is the (n - 10 + 1)th term from the beginning

= 20 - 10 + 1 = 11th term from beginning.

⇒ a_n = a + (n - 1)d

⇒ a₁₁ = 17 + (11 - 1)(-3)

= 17 + 10(-3)

= 17 - 30

= -13.

Assertion (A) is false.

The standard and correct formula for finding the nth term of an Arithmetic Progression.

a_n = a + (n - 1)d

Reason (R) is true.

A is false, R is true

Hence, option 2 is the correct option.

Assertion (A): For an A.P., T₂₂ = 149 and d = 7. Then S₂₂ is 1661.

Reason (R): The sum of first n terms of an A.P. is given by S_n = $\dfrac{n}{2}[2a − (n − 1)d]$.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

Given,

a₂₂ = 149

n = 22

d = 7.

We know that,

⇒ a_n = a + (n - 1)d

⇒ a₂₂ = a + (22 - 1)7

⇒ 149 = a + (21)7

⇒ 149 = a + 147

⇒ 149 - 147 = a

⇒ a = 2.

We know that,

S_n = $\dfrac{n}{2}$ (a + l)

⇒ S₂₂ = $\dfrac{22}{2}$ (2 + 149)

= 11 × (151)

= 1661.

Assertion (A) is true.

The standard and correct formula for the sum of the first n terms of an A.P.

S_n = $\dfrac{n}{2}$ [2a + (n − 1)d]

Reason (R) is false.

A is true, R is false

Hence, option 1 is the correct option.

Assertion (A): If the sum of first n terms of an A.P. is given S_n = 2n² − n, then its nth term is 4n + 3.

Reason (R): The nth term (t_n) of an A.P. is given by S_n − S_{n + 1}.

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false

Answer

We know that,

T_n = S_n - S_{n - 1}

Given,

S_n = 2n² - n

S_{n - 1} = 2(n - 1)² - (n - 1)

= 2(n² - 2n + 1) - n + 1

= 2n² - 4n + 2 - n + 1

= 2n² - 5n + 3.

T_n = 2n² - n - (2n² - 5n + 3)

= 2n² - n - 2n² + 5n - 3

= 4n - 3.

Assertion (A) is false.

The nth term of an A.P. is given by,

T_n = S_n - S_{n - 1} is correct.

Reason (R) is false.

Both A and R are false.

Hence, option 4 is the correct option.