If the n^{th} term of the geometric progressi | vedclass

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(A) The $n^{th}$ term of a geometric progression is given by $T_n = ar^{n-1}$.Here,the first term $a = 5$ and the common ratio $r = \frac{-5/2}{5} = -

Vedclass · Accepted Answer

$11$

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(A) The $n^{th}$ term of a geometric progression is given by $T_n = ar^{n-1}$.Here,the first term $a = 5$ and the common ratio $r = \frac{-5/2}{5} = -\frac{1}{2}$.Given $T_n = \frac{5}{1024}$,we have:$\frac{5}{1024} = 5 	imes (-\frac{1}{2})^{n-1}$Dividing both sides by $5$:$\frac{1}{1024} = (-\frac{1}{2})^{n-1}$Since $1024 = 2^{10}$,we can write:$(-\frac{1}{2})^{10} = (-\frac{1}{2})^{n-1}$Comparing the exponents:$10 = n - 1$$n = 11$.

If the $n^{th}$ term of the geometric progression $5, - \frac{5}{2}, \frac{5}{4}, - \frac{5}{8}, \dots$ is $\frac{5}{1024}$,then the value of $n$ is:

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