Question

Find the 10th term of the sequence 1000, -500, 250, -125, ...

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Sequence is 1000, -500, 250, -125, .  . . . . .}$

$\underline{\textbf{To find:}}$

$\textsf{10 th term of the given sequence}$

$\underline{\textbf{Solution:}}$

$\underline{\textbf{Concept used:}}$

$\mathsf{The\;n\;th\;term\;of\;the\;G.P\;a,ar,ar^2,\;.\;.\;.\;.\;.is}$

$\boxed{\mathsf{t_n=a\,r^{n-1}}}$

$\textsf{Clearly, the Sequence is 1000, -500, 250, -125, .  . . . . .is G.P}$

$\mathsf{Common\;ratio\;(r)=\dfrac{t_2}{t_1}=\dfrac{-500}{1000}}$

$\implies\mathsf{r=\dfrac{-1}{2}}$

$\mathsf{Also,\;a=1000}$

$\textbf{10 th term of the sequence}$

$\mathsf{t_{10}=a\,r^{9}}$

$\mathsf{t_{10}=1000\left(\dfrac{-1}{2}\right)^{9}}$

$\mathsf{t_{10}=1000\left(\dfrac{-1}{2^9}\right)}$

$\mathsf{t_{10}=125{\times}8\left(\dfrac{-1}{2^6{\times}2^3}\right)}$

$\mathsf{t_{10}=125\left(\dfrac{-1}{2^6}\right)}$

$\implies\boxed{\mathsf{t_{10}=\dfrac{-125}{64}}}$

$\therefore\mathsf{10\;th\;term\;of\;the\;sequence\;is\;\dfrac{-125}{64}}$

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