good one

The sum of the first few terms in a geometric progression is 11, the sum of their squares is 341, and sum of their cubes is 3641. Find the terms of the sequence.

3 Answers

24
eureka123 ·

S=a+ar+ar2....+arn-1=a(1-rn)1-r

S1=a2+a2r2+a2r4+...+a2r2(n-1)=a2(1-r2n)1-r2

S2=a3+a3r3+a3r6+...+a3r3(n-1)=a3(1-r3n)1-r3

Dividing them and some rearrangement will get the answer..

1
Arshad ~Died~ ·

can anyone please post full solution...i have some problem with this one....

21
eragon24 _Retired ·

@arshad wat prob r you having with it......u hav 3 eq... 3 unknowns....solve it

Your Answer

Close [X]