24S=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..
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.
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3 Answers
24
1can anyone please post full solution...i have some problem with this one....
21@arshad wat prob r you having with it......u hav 3 eq... 3 unknowns....solve it