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(C) Let the first term be $a$ and the common ratio be $r$. The terms are $a, ar, ar^2, ar^3, \dots$Given that the third term $a_3 = a_1^2$,we have $ar

Vedclass · Accepted Answer

$128$

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(C) Let the first term be $a$ and the common ratio be $r$. The terms are $a, ar, ar^2, ar^3, \dots$Given that the third term $a_3 = a_1^2$,we have $ar^2 = a^2$.Since $a 
eq 0$,we can divide by $a$ to get $r^2 = a$.Given the second term $a_2 = ar = 8$,we substitute $a = r^2$ into this equation:$r^2 \cdot r = 8 \implies r^3 = 8 \implies r = 2$.Now,find $a$: $a = r^2 = 2^2 = 4$.The sixth term is given by $a_6 = ar^5$.$a_6 = 4 	imes (2)^5 = 4 	imes 32 = 128$.

The third term of a geometric progression is equal to the square of the first term. If its second term is $8$,then its sixth term will be:

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