Question

How many terms of the geometric progression $1+4+16+64+..........$ must be added to get sum equal to $5641$?

Answer 1

$GP=4^0,4^1,4^2,4^3......4^n$ $(n+1)$ terms

Sum $=5641$

Sum of $GP=\frac{a(r^{n+1}-1)}{r-1}$

$=\frac{1(4^{n+1}-1)}{4-1}=5641$

$\Rightarrow 4^{n+1}=16924$

$n+1={log}_{4}16924$

$=7.023\approx 7$ terms.

Sum of $7$ terms $=5461$.