If a , b , c are in G P and a 1 x = b 1 y = c 1 2 then prove that x , y , z are in A P .

It is given that: a 1 x = b 1 y = c 1 2 Let a 1 x = b 1 y = c 1 2 = k ⇒ a 1 x = k ⇒ a 1 x x = k x … (Taking power of x on both sides.) ⇒ a 1 x × x = k x ⇒ a = k x Similarly b = k y And c = k 2 It is given that a , b , c are in G . P . ⇒ b 2 = a c Substituting values of a , b , c into above equation, we get: ⇒ k y 2 = k x k z ⇒ k 2 y = k x + z Comparing the powers we get, 2 y = x + z y = x + z 2 y is the arithmean mean of x and z ⇒ x , y , z , are in A . P .

Express 0 . 123 ¯ as a rational number.

Let, x = 0 . 123123123 … ⇒ x = 0 . 123 + 0 . 000123 + 0 . 000000123 + … . ∞ ⇒ x = 123 ( 0 . 001 + 0 . 000001 + 0 . 000000001 + … . ∞ ) ⇒ x = 123 1 10 3 + 1 10 6 + 1 10 9 + 1 10 12 + … ∞ Consider 1 10 3 + 1 10 6 + 1 10 9 + 1 10 12 + … ∞ Here, the first term a = 1 10 3 and common ratio r = 1 10 3 This is an infinite geometric series We know that sum of infinite geometric series is = a 1 - r ∴ Sum = 1 10 3 1 - 1 10 3 = 1 × 1000 999 × 1000 = 1 999 ⇒ x = 123 × 1 999 = 123 999 Ans : 0 . 123 ¯ = 123 999

Express 0 . 6 ¯ as a rational number.

Let x = 0 . 6666 … ⇒ x = 0 . 6 + 0 . 06 + 0 . 006 + … ⇒ x = 6 ( 0 . 1 + 0 . 01 + 0 . 001 + 0 . 0001 + … . ∞ ) ⇒ x = 6 1 10 + 1 100 + 1 1000 + 1 10000 + … ∞ Consider 1 10 + 1 100 + 1 1000 + 1 10000 + … ∞ Here, a = 1 10 and r = 1 10 This is an infinite geometric series. We know that the sum of infinite terms of a GP S ∞ = a 1 - r ∴ S ∞ = a 1 - r = 1 10 1 - 1 10 = 1 × 10 9 × 10 = 1 9 ∴ x = 6 × 1 9 = 6 9 = 2 3 Ans: 0 . 6 ¯ = 2 3 .

Express 0 . 6 8 ¯ as a rational number.

Let, x = 0 . 68888888 … ⇒ 10 x = 6 . 88888 . . . . . . ⇒ 10 x = 6 + 0 . 8 + 0 . 08 + 0 . 008 + … ∞ ⇒ x = 6 + 8 ( 0 . 1 + 0 . 01 + … ∞ ) ⇒ x = 6 + 8 1 10 + 1 10 2 + 1 10 3 + … ∞ Consider, 1 10 + 1 10 2 + 1 10 3 + … ∞ Here, the first term a = 1 10 and common ratio r = 1 10 This is an infinite geometric series. We know that sum of infinite geometric series S ∞ = a 1 - r ∴ Sum = a 1 - r = 1 10 1 - 1 10 = 1 × 10 9 × 10 = 1 9 10 x = 6 + 8 × 1 9 = 54 + 8 9 = 62 9 ⇒ x = 62 90 Ans: 0 . 68 ¯ = 62 90 = 31 45 .

The sum of first three terms of a GP is 39 10 and their product is 1 . Find the common ratio and these three terms.

Let the first three terms of G.P. be a r , a , a r It is given that a r × a × a r = 1 ⇒ a 3 = 1 ⇒ a = 1 And a r + a + a r = 39 10 ⇒ a 1 r + 1 + r = 39 10 ⇒ 1 r + 1 + r = 39 10 … ( a = 1 ) ⇒ 1 r + r = 39 10 - 1 = 29 10 ⇒ 10 1 + r 2 = 29 r ⇒ 10 r 2 - 29 r + 10 = 0 ⇒ 10 r 2 - 25 r - 4 r + 10 = 0 ⇒ 5 r ( 2 r - 5 ) - 2 ( 2 r - 5 ) = 0 ⇒ ( 2 r - 5 ) ( 5 r - 2 ) = 0 ⇒ r = 5 2 · 2 5 Therefore the first three terms are: i ) If r = 5 2 then 2 5 , 1 , 5 2 ii ) If r = 2 5 then the terms are 5 2 , 1 , 2 5

E M B I B E

Senior Secondary School Mathematics FOR CLASS 11>Algebra>Chapter 12 - Geometrical Progression>EXERCISE 12H

R S Aggarwal Solutions for Exercise 8: EXERCISE 12H

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 8: EXERCISE 12H

Attempt the practice questions from Exercise 8: EXERCISE 12H with hints and solutions to strengthen your understanding. Senior Secondary School Mathematics FOR CLASS 11 solutions are prepared by Experienced Embibe Experts.