# In a G.P , sum of the first three terms are 13.Sum of the square of the first three terms are 91 . Find the G.P.

1
by nandhana

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by nandhana

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a/q,

(a/x)^2+a+(ax)^2 = 91 ------------eqn(1)

a/x +a+ax = 13 --------eqn(2)

or, a(x^2+x+1) = 13x

we know that ;

(a/x+a+ax)^2 = (a/x)^2+a^2+(ax)^2+2a^2/x+2a^2x+2a^2

now put the values of (a/x+a+ax) & (a/x)^2+a^2+(ax)^2 from eqn (1) & (2).

or, 13^2 = 91+2a^2(1/x+x+1)

or, 2a^2(x^2+x+1)/x = 169-91

or, 2a^2(x^2+x+1) = 68x

put the value of x from (1)

or, 2a^2(x^2+x+1) = 68a(x^2+x+1)/13

or, 2a=68/13

or, a=34/13

now we can find the value of x,when put the value of a in (2)

so series of GP 34/13x, 34/13, 34x/13