Arithmetic Progression — Exercise 10(A)

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) 16th 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.

⇒ a₁₆ = 13 + (16 - 1)7

= 13 + (15) × 7

= 13 + 105

= 118.

Hence, a = 13, d = 7 and a₁₆ = 118.

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°.

-11, -7, -3, .......,49, 53 are the terms of a progression.

Answer the following:

(i) What is the type of progression?

(ii) How many terms are there in all?

(iii) Calculate the value of middle most term.

Answer

(i) Given,

-11, -7, -3, ......,49, 53

Second term - First term : −7 − (−11) = 4

Third term - Second term : −3 − (−7) = 4

Last term - Second last term : 53 − 49 = 4

Since the difference between consecutive terms is constant, the progression is an Arithmetic Progression (AP) with first term (a) = -11 and common difference (d) = 4.

Hence, the progression is an Arithmetic Progression.

(ii) Let 53 be the nth term.

By formula,

⇒ t_n = a + (n - 1)d

⇒ 53 = -11 + (n - 1)4

⇒ 53 = -11 + 4n - 4

⇒ 53 = -15 + 4n

⇒ 4n = 53 + 15

⇒ 4n = 68

⇒ n = $\dfrac{68}{4}$

⇒ n = 17.

Hence, there are a total of 17 terms in the progression.

(iii) Since, n = 17, is odd.

By formula,

Middle term = $\dfrac{n + 1}{2}$

= $\dfrac{17 + 1}{2}$

= $\dfrac{18}{2}$

= 9.

Thus, 9th term is middle term.

By formula,

⇒ t₉ = a + (n - 1)d

= -11 + (9 - 1) × 4

= -11 + 8 × 4

= -11 + 32

= 21.

Hence, the value of the middle most term is 21.