Question

Sum of the first three terms of a gp is 13 as sum of their squares 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.