Question

The sum of an infinite geometric progression is 15 and the sum of the squares of these terms is 45. Find the G.P.

Answer 1

a/c to question,

The sum of an infinite geometric progression is 15.
Let a is first term and r is the common ratio of geometric progression then,
$S_{\infty}=\frac{a}{1-r}$

so, a/(1 - r) = 15

or, a²/(1 - r)² = 225 ......(1)

again, sum of square of given terms in gp = 45
if a, ar , ar², ar³, ar⁴ .... are terms in gp
then, a², a²r², a²r⁴ ....... are square of those terms in gp.

so, a²/(1 - r²) = 45

or, a²/(1 - r)(1 + r) = 45 .....(2)


from equations (1) and (2),

(1 + r)/(1 - r) = 225/45 = 5

or, 1 + r = 5 - 5r

or, 6r = 5 - 1 = 4

hence, r = 2/3 put it in equation (1) we get, a = 5

so, terms in GP are : 5, 10/3, 20/9 , 40/27 .....

Answer 2

Easy explanation infew