Question

Find 3 consecutive terms in GP whose sum is 13 and sum of whose square is 91?

Answer 1

Let the numbers be a , ar , ar² .

Now a + ar + ar² = 13

a(1 + r + r²) = 13 ---------(1)

a² + a²r² + a²r⁴ = 91

a² ( 1 + r² + r⁴ ) = 91 --------(2)

Squaring (1) dividing by (2)

a²(1 + r + r²)² / a² ( 1 + r² + r⁴ ) = 169 / 91.

(1 + r + r²)² / ( 1 + r² )² – r² ) = 13 / 7.

(1 + r + r²)² / ( 1 + r² + r) ( 1 + r² - r) = 13 / 7.

( 1 + r² + r) / ( 1 + r² - r) = 13 / 7

7( 1 + r² + r) = 13( 1 + r² - r)

( 7 + 7r² + 7r) = ( 13 + 13r² - 13 r)

6r² - 20r + 6 = 0

3r² - 10r + 3 = 0.

3r² - 9r - r+ 3 = 0.

3r(r – 3) –1 (r - 3) = 0.

(3r – 1)(r – 3) = 0

r = 3 , 1 / 3.

Substitute r in equ(1) we get

a( 1 + 3 + 9) = 13 and a ( 1 + 1 / 3 + 1 /9) = 13

13a = 13 and 13a / 9 = 13

a = 1 and a = 9.

Now numbers are a , ar , ar².

If r = 3 and a = 1 then

1, 3 , 9 are the numbers.

If r = 1/ 3 and a = 9 then

9, 3 , 1 are the numbers.