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Geometric Progression — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsGeometric Progression1 Mark
MCQ
Which term of the G.P., $18,-12,8, \ldots$ is $\frac{512}{729} ?$
  • A
    $12^{\text {th }}$
  • B
    $11^{\text {th }}$
  • C
    $10^{\text {th }}$
  • ✓
    $9^{\text {th }}$
✓

Answer

Correct option: D.
$9^{\text {th }}$
(d) $9^{\text {th }}$
Explanation :
Which term of the G.P. $18,-12,8, \ldots \frac{512}{729}$
First term $(a)=18$,
Common ratio $(r)=\frac{-12}{18}-\frac{-2}{3}$
$\begin{array}{l}\therefore a_n=a r^{n-1} \\ \frac{512}{729}=18\left(\frac{-2}{3}\right)^{n-1} \\ \frac{512}{729} \times \frac{1}{18}=\left(\frac{-2}{3}\right)^{n-1} \\ \frac{256}{729 \times 9}=\left(\frac{-2}{3}\right)^{n-1} \\ \left(\frac{-2}{3}\right)^8=\left(\frac{-2}{3}\right)^{n-1}\end{array}$
Comparing $n-1=8$
n=8+1
n=9
It is $9^{\text {th }}$ term.

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