The sum of a few terms of a geometric progres | vedclass

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(A) Let the first term be $a$,common ratio $r = 3$,and the number of terms be $n$.The $n^{th}$ term of a geometric progression is given by $T_n = a r^

Vedclass · Accepted Answer

$2$

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(A) Let the first term be $a$,common ratio $r = 3$,and the number of terms be $n$.The $n^{th}$ term of a geometric progression is given by $T_n = a r^{n-1}$.Given $T_n = 486$,we have $a(3)^{n-1} = 486$. Multiplying both sides by $3$,we get $a(3)^n = 1458$ .....$(i)$.The sum of $n$ terms is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.Given $S_n = 728$ and $r = 3$,we have $728 = \frac{a(3^n - 1)}{3 - 1}$.$728 = \frac{a(3^n) - a}{2}$.$1456 = a(3^n) - a$ .....$(ii)$.Substituting the value of $a(3^n)$ from equation $(i)$ into equation $(ii)$:$1456 = 1458 - a$.$a = 1458 - 1456 = 2$.Thus,the first term is $2$.

The sum of a few terms of a geometric progression is $728$. If the common ratio is $3$ and the last term is $486$,then the first term of the series will be:

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