Gujarat BoardEnglish MediumSTD 11 CommerceStatisticsGeometric progression2 Marks
Question
For a $G.P.a =\frac{4}{9}$ and $r =-\frac{3}{2}$. Find $T_3$.
✓
Answer
$a=\frac{4}{9^{\prime}} r=-\frac{3}{2^2} T_3=?$
$T_n=a \cdot r^{n-1}$
$\therefore T_3=\left(\frac{4}{9}\right)\left(-\frac{3}{2}\right)^{3-1}=\left(\frac{4}{9}\right)\left(-\frac{3}{2}\right)^2=\frac{4}{9} \times \frac{9}{4}=1$
Hence, the third term of a $G.P.$ is $1.$
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