AP Inter 1st Year Maths Exercise 8d Solutions

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8d

Find the sum to infinity in each of the following Geometric Progression.

Question 1.
1, \(\frac{1}{3}, \frac{1}{9}\), …………..
Solution:
Let S = 1, \(\frac{1}{3}, \frac{1}{9}\), …………..; Here, a = 1 and r = \(\frac{1}{3}\)
So, S_∞ = \(\frac{a}{1-r}=\frac{1}{1-\frac{1}{3}}=\frac{1}{2 / 3}=\frac{3}{2}\) = 1.5

Question 2.
6, 1.2, 0.24, …………………
Solution:
Let S = 6 + 1.2 + 0.24 + ………………….; Here, a = 6 and r = 0.2
So, S_∞ = \(\frac{a}{1-r}=\frac{6}{1-(0.2)}=\frac{6}{0.8}\) = 7.5

Question 3.
5, \(\frac{20}{7}, \frac{80}{49}\) ………..
Solution:
Let S = 5, \(\frac{20}{7}, \frac{80}{49}\), ………….; Here a = 5, r = \(\frac{4}{7}\)
So, S_∞ = \(\frac{a}{1-r}=\frac{5}{1-\frac{4}{7}}=\frac{35}{3}\)

Question 4.
\(\frac{-3}{4}, \frac{3}{16}, \frac{-3}{64}\) ………….
Solution:
Let S = \(\frac{-3}{4}, \frac{3}{16}, \frac{-3}{64}\) ………….; Here, a = \(\frac{-3}{4}\) and r = \(\frac{-1}{4}\)

S_∞ = \(\frac{a}{1-r}=\frac{-3 / 4}{1-\left(\frac{-1}{4}\right)}=\frac{-3 / 4}{1+\frac{1}{4}}=\frac{-3}{4} \times \frac{4}{5}=\frac{-3}{5}\)

Question 5.
Prove that 3^{\(\frac{1}{2}\)} × 3^{\(\frac{1}{4}\)} × 3^{\(\frac{1}{8}\)} × ……. = 3
Solution:
LHS = 3^{\(\frac{1}{2}\)} × 3^{\(\frac{1}{4}\)} × 3^{\(\frac{1}{8}\)} × ……. = 3^{\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots .\)};
Here a = \(\frac{1}{2}\) and r = \(\frac{1}{2}\)
∴ S_∞ = 3^{\(\left(\frac{a}{1-r}\right)\)}
= 3^{\(\left(\frac{1 / 2}{1-1 / 2}\right)\)}
= 3^{\(\left(\frac{1 / 2}{1 / 2}\right)\)}
= 3¹
= 3 = R.H.S. [using S_∞ = \(\frac{a}{1-r}\)]

Question 6.
Let x = 1 + a + a² + …….. and y = 1 + b + b² + ….. where |a| < 1 and |b| < 1.
Prove that 1 + ab + a²b² + ….. = \(\frac{x y}{x+y-1}\)
Solution:
Here, x = 1 + a + a² + ……………… = \(\frac{1}{1-a}\) [using S_∞ = \(\frac{a}{1-r}\)]

And y = 1 + b + b² + ………… = \(\frac{1}{1-b}\)

LHS = 1 + ab + a²b² + ……. = \(\frac{1}{1-ab}\) …………(1)

RHS = \(\frac{x y}{x+y-1}=\frac{\left(\frac{1}{1-a}\right)\left(\frac{1}{1-b}\right)}{\left(\frac{1}{1-a}\right)+\left(\frac{1}{1-b}\right)-1}=\frac{\frac{1}{(1-a)(1-b)}}{\frac{1-b+1-a-(1-a)(1-b)}{(1-a)(1-b)}}\)

= \(\frac{1}{(1-a)(1-b)} \times \frac{(1-a)(1-b)}{2-a-b-1+a+b-a b}=\frac{1}{1-a b}\) ……. (2)

From (1) and (2), 1 + ab + a²b² + ………… = \(\frac{x y}{x+y-1}\)