Q. 115.0( 1 Vote )

# The sum of the first three terms of a G.P. is 13 and sum of their squares is 91. Determine the G.P.

Let the first term of G.P be a

second term = ar and third term = ar2.

(where, r is the common ratio)

sum of three terms is 13

a(1 + r + r2) = 13r ……..(1)

Also, sum of their squares is 91.

a2 (1 + r2 + r4) = 91r2 ………(2)

Now, Squaring (1) dividing by (2)    7( 1 + r2 + r) = 13( 1 + r2 - r)

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

6r2 - 20r + 6 = 0

2(3r2 - 10r + 3) = 0

3r2 - 10r + 3 = 0

3r2 - 9r-r + 3 = 0

(3r – 1)(r – 3) = 0 Substituting r in equation (1), we get-

a(1 + 3 + 9) = 13 × 3

And, 13a = 13

And a = 1 and a = 9.

Now, G.P is-

a, ar, ar2,….

If r = 3 and a = 1 then,

1, 1×3, 1×32, ……

= 1, 3 , 9 are the first three terms.

And, If and a = 9 then, = 9, 3, 1 are the first three terms.

1,3,9,… … or 9,3,1,… … is the G.P.

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