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.

Answer :

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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