Find the 9^{th} term of the geometric progres | vedclass

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(A) Given the geometric progression $8, 12, 18, 27, \dots$Here,the first term $a = 8$ and the common ratio $r = \frac{12}{8} = \frac{3}{2}$.The $n^{th

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$6561/32$

Vedclass · Answer

(A) Given the geometric progression $8, 12, 18, 27, \dots$Here,the first term $a = 8$ and the common ratio $r = \frac{12}{8} = \frac{3}{2}$.The $n^{th}$ term of a geometric progression is given by $T_n = a \cdot r^{n-1}$.For the $9^{th}$ term $(n = 9)$:$T_9 = 8 \cdot (\frac{3}{2})^{9-1}$$T_9 = 8 \cdot (\frac{3}{2})^8$$T_9 = 8 \cdot \frac{6561}{256}$$T_9 = \frac{6561}{32}$

Find the $9^{th}$ term of the geometric progression $8 + 12 + 18 + 27 + \dots$

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