Arithmetic Progression — Exercise 10(B)

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 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,

∴ an = a + (n - 1)d

Given,

a = 101

an = 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)

⇒ $\dfrac{25}{2}$ ×(250)

= 25 × 125

= 3125.

Hence, S₂₅ = 3125.

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.

The nth term of an Arithmetic Progression (A.P.) is given by the relation T_n = 6(7 - n). Find ∶

(i) its first term and common difference

(ii) sum of its first 25 terms

Answer

T_n = 6(7 - n)

Substituting n = 1, we get :

T₁ = 6(7 - 1) = 6 × 6 = 36.

Substituting n = 2, we get :

T₂ = 6(7 - 2) = 6 × 5 = 30.

Common difference (d) = T₂ - T₁

= 30 - 36 = -6.

Hence, common difference (d) = -6 and first term = 36.

(b) By formula,

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

= $\dfrac{25}{2}$ [2 × 36 + (25 − 1) × −6]

= 12.5 × [72 + 24× − 6]

= 12.5 × [72 − 144]

= 12.5 × −72

= −900.

Hence, sum of its first 25 terms = -900.

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.

164, 160, 156, 152, ..... are in Arithmetic Progression (A.P.). Find :

(i) which term is equal to 0.

(ii) the sum of its first 20 terms.

Answer

Given,

First term (a) = 164

Common difference (d) = 160 - 164 = -4

(i) Let n^th term be zero.

⇒ a_n = 0

⇒ a + (n - 1)d = 0

⇒ 164 + (n - 1)(-4) = 0

⇒ 164 - 4n + 4 = 0

⇒ 168 - 4n = 0

⇒ 4n = 168

⇒ n = $\dfrac{168}{4}$

⇒ n = 42.

Hence, 42nd term is equal to 0.

(ii) By formula,

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

Substituting values we get :

$S_{20} = \dfrac{20}{2}\Big[2 \times (164) + (20 - 1) \times (-4)\Big] \\[1em] \Rightarrow S_{20} = 10[328 + (19) \times (-4)] \\[1em] \Rightarrow S_{20} = 10[328 - 76] \\[1em] \Rightarrow S_{20} = 10 \times 252 \\[1em] \Rightarrow S_{20} = 2520.$

Hence, sum of first 20 terms = 2520.

An arithmetic progression (A.P.) has 3 as its first term. The sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference of the A.P.

Answer

Let common difference be d.

a = 3

Sum of first n terms of an A.P. = $\dfrac{\text{n}}{2}(\text{a } + \text{ l})$

Given,

The sum of the first 8 terms is twice the sum of the first 5 terms.

$\therefore \dfrac{8}{2}(a + a_8) = 2 \times \dfrac{5}{2}(a + a_5) \\[1em] \Rightarrow 4[a + a + (8 - 1)d] = 5[a + a + (5 - 1)d] \\[1em] \Rightarrow 4[2a + 7d] = 5[2a + 4d] \\[1em] \Rightarrow 4[2 \times 3 + 7d] = 5[2 \times 3 + 4d] \\[1em] \Rightarrow 4[6 + 7d] = 5[6 + 4d] \\[1em] \Rightarrow 24 + 28d = 30 + 20d \\[1em] \Rightarrow 28d - 20d = 30 - 24 \\[1em] \Rightarrow 8d = 6 \\[1em] \Rightarrow d = \dfrac{6}{8} = \dfrac{3}{4}.$

Hence, common difference = $\dfrac{3}{4}$.

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 in the flower bed? how many rose plants are there in the flower 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.