OP Malhotra Class 10 Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(d)

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S Chand Class 10 ICSE Maths Solutions Chapter 9 Arithmetic and Geometric Progression Ex 9(d)

Question 1.
Find the sum of the terms of the indicated geometric sequence.
(i) 3, -6, 12,… to 6 terms
(ii) -2, -6, -18,… to 7 terms
(iii) \(\frac { 1 }{ 9 }\), \(\frac { 1 }{ 3 }\), 1, … to 5 terms 1
(iv) 2, 1, \(\frac { 1 }{ 2 }\), … to 6 terms
(v) 1, y, … to 10 terms
(vi) 0.15, 0.015, 0.0015, … 20 terms
Solution:
Find the sum of
(i) 3, – 6, 12, … to 6 terms
Here, a = 3, r = \(\frac { -6 }{ 3 }\) = – 2, n = 6
S₆ = \(\frac{a\left(1-r^n\right)}{1-r}\) (∵ r < 1)
= \(\frac{3\left[1-(-2)^6\right]}{1-(-2)}=\frac{3(1-64)}{1+2}\)
= \(\frac { 3(-63) }{ 3 }\) = – 63

(ii) -2, -6, -18,… to 7 terms
Here, a = – 2, r = \(\frac { -6 }{ -2 }\) = 3, n = 7
∴ S₇ = \(\frac{a\left(r^n-1\right)}{r-1}\) (∵ r > 1)
= \(\frac{-2\left(3^7-1\right)}{3-1}=\frac{-2(2187-1)}{2}\)
= – (2186) = – 2186

(iii) \(\frac { 1 }{ 9 }\), \(\frac { 1 }{ 3 }\), 1, … to 5 terms
Here, a = \(\frac { 1 }{ 9 }\) r = \(\frac { 1 }{ 3 }\) ÷ \(\frac { 1 }{ 9 }\) = \(\frac { 1 }{ 3 }\) x \(\frac { 9 }{ 1 }\) = 3,
S₅ = \(\frac{a\left(r^n-1\right)}{r-1}=\frac{\frac{1}{9}\left(3^5-1\right)}{3-1}\)
= \(\frac{1}{9 \times 2}(243-1)=\frac{1}{18} \times 242\)
= \(\frac { 121 }{ 9 }\) = 13\(\frac { 4 }{ 9 }\)

(iv) 2, 1, \(\frac { 1 }{ 2 }\), … to 6 terms
Here, a = 2, r = 1 ÷ 2 = \(\frac { 1 }{ 2 }\), n = 6

(v) 1, \(\frac { 2 }{ 3 }\), \(\frac { 4 }{ 9 }\), … to 10 terms

(vi) 0.15, 0.015, 0.0015, … 20 terms
Here, a = 0.15, r = \(\frac { 0.015 }{ 0.15 }\) = \(\frac { 1 }{ 10 }\) = 0.1, n = 20
∴ S₂₀ = \(\frac{a\left(1-r^n\right)}{1-r}=\frac{0.15\left[1-(0.1)^{20}\right]}{1-0.1}\)
= \(\frac{0.15}{0.9}\left[1+(0.1)^{20}\right]=\frac{15}{90}\left[1+(0.1)^{20}\right]\)
= \(\frac { 1 }{ 6 }\)[1 + (0.1)²⁰]

Question 2.
Find the sum of the following series
(i) 12 + 6 + 3 + 1.5 + … to 10 terms
(ii) 6 – 3 + 1\(\frac { 1 }{ 2 }\) – \(\frac { 3 }{ 4 }\) + … to 15 terms
(iii) 2 + 6 + 18 + 54 + … to 12 terms
(iv) 6 + 12 + 24 + … + 1536
Solution:
(i) 12 + 6 + 3 + 1.5 + … to 10 terms
Here, a = 12, r = \(\frac { 6 }{ 12 }\) = \(\frac { 1 }{ 2 }\), n = 10

(ii) 6 – 3 + 1\(\frac { 1 }{ 2 }\) – \(\frac { 3 }{ 4 }\)+… to 15 terms
Here, a = 6, r = \(\frac { -3 }{ 6 }\) = \(\frac { -1 }{ 2 }\), n = 15

(iii) 2 + 6 + 18 + 54 + … to 12 terms 6
Here, a = 2, r = \(\frac { 6 }{ 2 }\) = 3, n = 12
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\) (∵ r > 1)
S₁₂ = \(\frac{2\left(3^{12}-1\right)}{3-1}=\frac{2\left(3^{12}-1\right)}{2}\) = 3¹² – 1

(iv) 6 + 12 + 24 + … + 1536 12
Here, a = 6, r = \(\frac { 12 }{ 6 }\) = 2, l = 1536
T_n = l = a(rⁿ – 1)
1536 = 6(2^n-1)

Question 3.
Find the sum to n terms of the following series.
(i) 12 + 6 + 3 + 1 \(\frac { 1 }{ 2 }\) + …
(ii) 20 – 10 + 5 – 2\(\frac { 1 }{ 2 }\) + …
(iii) 9 – 3, +1 – \(\frac { 1 }{ 3 }\) + …
(iv) \(\sqrt{3}\) + 3 + 3\(\sqrt{3}\) + 9 + …
(v) 0.9 + 0.09 + 0.009 + 0.0009 + ….
Solution:
(i) 12 + 6 + 3 + 1 \(\frac { 1 }{ 2 }\) + … n terms

(ii) 20 – 10 + 5 – 2\(\frac { 1 }{ 2 }\) + … n terms
Here, a = 20, r = \(\frac { -10 }{ 20 }\) = \(\frac { -1 }{ 2 }\)

(iii) 9 – 3, + 1 – \(\frac { 1 }{ 3 }\)+… n terms

(iv) \(\sqrt{3}\) + 3 + 3\(\sqrt{3}\) + 9 + … n terms
Here, a = \(\sqrt{3}\), r = \(\frac{3}{\sqrt{3}}\) = \(\sqrt{3}\)

(v) 0.9 + 0.09 + 0.009 + 0.0009 + ….
Here, a = 0.9, r = \(\frac { 0.09 }{ 0.9 }\) = \(\frac { 1 }{ 10 }\) = 0.1
S_n = \(\frac{a\left(1-r^n\right)}{1-r}=\frac{0.9\left[1-(0.1)^n\right]}{1-0.1}\)
= \(\frac { 0.9 }{ 0.9 }\)[1- (0.1)ⁿ]
= 1 – (0.1)ⁿ

Question 4.
Use the formula S_n = \(\frac{a\left(r^n-1\right)}{r-1}\) to find:
(i) a₁ and S_n if a_n = 1000, r = 10 and n = 7
(ii) n and S if a_n = 5, a₁ = 320, r = 2
(iii) a₉ and a₁ if n = 9, r = 2, s_n = 1022
Solution:
S_n = \(\frac{a\left(r^{n^{\prime}}-1\right)}{r-1}\)
(i) a_n = 1000, r = 10 and n = 7, a₁ and S_n
a_n = ar^n-1 ⇒ a x 10^7-1 = 1000

(ii) a₁ = 5, a_n = 320, r = 2, n and S_n
a_n = ar^n-1 ⇒ 320 = 5 × 2^2n-1
2^n-1 = \(\frac { 320 }{ 5 }\) = 64 = 2⁶
∴ n – 1 = 6 ⇒ n = 6 + 1 = 7
and S_n = \(\frac{a\left(r^n-1\right)}{r-1}=\frac{5\left(2^7-1\right)}{2-1}\)
= \(\frac{5(128-1)}{1}\) = 5 × 127 = 635

(iii) n = 9, r = 2, S_n = 1022, a₉ and a₁

Question 5.
If {a_n} is a G.S. (i.e., geometric sequence) and a₁ = 4, r = 5, find a₆ and S₆.
Solution:
{a_n} is in G.S.
a₁ = 4, r = 5, find a₆ and S₆,
S₆ = ar^n-1 = 4 x (5)^6-1 = 4 x 5⁵
= 4 x 5 x 5 x 5 x 5 x 5 = 12500
S₆ = \(\frac { 1 }{ 2 }\)
= 5⁶ – 1 = 15625 – 1 = 15624

Question 6.
How many terms of the G.P. 3, \(\frac { 3 }{ 2 }\), \(\frac { 3 }{ 4 }\),…are needed to give the sum
Solution:
G.P. is
3, \(\frac { 3 }{ 2 }\), \(\frac { 3 }{ 4 }\), …. sum = \(\frac { 3069 }{ 512 }\)

Question 7.
The sum of some terms of a G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.
Solution:
Sum of some terms = 315
Ratio in first term (a₁) and common ratio (r) = 5:2
Let number of terms be n
and first term (a₁) be 5 and r = 2
Now, S_n = \(\frac{a\left(r^n-1\right)}{r-1}\)
315 = \(
63 = 2^{n – 1} ⇒ 2ⁿ = 63 + 1 = 64 = 2⁶
∴ n = 6
and T₆ or a₆ = a^{n -1} = 5(2)^6-1
= 5 x 2⁵ = 5 x 32 = 160

Question 8.
Given a GP. with a = 729 and 7th term = 64, determine S₇.
Solution:
In a G.P.
a = 729, T₇ = 64, S₇ T₇ = ar^n-1
729.r^7-1 = 64 ⇒ 729.r⁶ = 64

Question 9.
Find the sum of the series 2 + 6 + 18 + … + 4374.
Solution:
G.P. is
2 + 6 + 18 + … + 4374 6
Here, a = 2, r = [latex]\frac { 6 }{ 2 }\) = 3, and l = 4374
Now, a₇ = l = ar^n-1
⇒ 4374 = 2 x 3^n-1 ⇒ 3^n-1 = \(\frac { 4374 }{ 2 }\)
⇒ 3^n-1 = 2187 = 3⁷
Comparing, we get
n – 1 = 7 ⇒ n = 7 + 1 = 8
∴ n = 8
Now, S₈ = \(\frac{a\left(r^n-1\right)}{r-1}=\frac{2\left(3^8-1\right)}{3-1}\)
= \(\frac{2(6561-1)}{2}\) = 6560

Question 10.
How many terms of the sequence \(\sqrt{3}\), 3, 3\(\sqrt{3}\), … must be taken to make the sum 39 + 13\(\sqrt{3}\)?
Solution:
G.P. is
\(\sqrt{3}\), 3, 3\(\sqrt{3}\), ….. and S_n = 39 + 13\(\sqrt{3}\)
Here, a = \(\sqrt{3}\), r = \(\sqrt{3}\)
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\) (∵ r > 1)
39 + 13 \(\sqrt{3}=\frac{\sqrt{3}\left[(\sqrt{3})^n-1\right]}{\sqrt{3}-1}\)
⇒ \((\sqrt{3})^n-1=\frac{(\sqrt{3}-1)(39+13 \sqrt{3})}{\sqrt{3}}\)
= (13\(\sqrt{3}\) + (\(\sqrt{3}\) – 1)
= (13\(\sqrt{3}\) + 1) (\(\sqrt{3}\) – 1)
= 13(\(\sqrt{3}\) – 1) = 13 x 2 = 26
∴ (\(\sqrt{3}\))ⁿ = 26 + 1 = 27 = 3³ = (\(\sqrt{3}\))⁶
Comparing, we get n = 6
Number of terms = 6
Comparing, we get
n = 6
∴ Number of terms = 6