The sum of the first 20 terms of a GP is 1025 | vedclass

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(B) The sum of the first $n$ terms of a $GP$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Given,$S_{20} = 1025 	imes S_{10}$.Substituting the formula

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$2$

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(B) The sum of the first $n$ terms of a $GP$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Given,$S_{20} = 1025 	imes S_{10}$.Substituting the formula: $\frac{a(r^{20} - 1)}{r - 1} = 1025 	imes \frac{a(r^{10} - 1)}{r - 1}$.Assuming $r 
eq 1$ and $a 
eq 0$,we simplify: $r^{20} - 1 = 1025(r^{10} - 1)$.Since $r^{20} - 1 = (r^{10})^2 - 1^2 = (r^{10} - 1)(r^{10} + 1)$,we have:$(r^{10} - 1)(r^{10} + 1) = 1025(r^{10} - 1)$.If $r^{10} - 1 
eq 0$,then $r^{10} + 1 = 1025$.$r^{10} = 1024$.$r^{10} = 2^{10}$.Therefore,$r = \pm 2$.

The sum of the first $20$ terms of a $GP$ is $1025$ times the sum of its first $10$ terms. Find the common ratio of the $GP$.

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