The sum of some terms of a geometric progress | vedclass

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(A) The $n^{th}$ term of the geometric progression is given by $a r^{n-1} = 486$. Given $r = 3$,we have $a(3)^{n-1} = 486$,which implies $a \cdot 3^n

Vedclass · Accepted Answer

$2$

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(A) The $n^{th}$ term of the geometric progression is given by $a r^{n-1} = 486$. Given $r = 3$,we have $a(3)^{n-1} = 486$,which implies $a \cdot 3^n = 3 	imes 486 = 1458$ (Equation $i$). The sum of $n$ terms is $S_n = \frac{a(r^n - 1)}{r - 1} = 728$. Substituting $r = 3$,we get $\frac{a(3^n - 1)}{3 - 1} = 728$,which simplifies to $a \cdot 3^n - a = 728 	imes 2 = 1456$ (Equation $ii$). Substituting the value of $a \cdot 3^n$ from Equation $i$ into Equation $ii$: $1458 - a = 1456$. Therefore,$a = 1458 - 1456 = 2$.

The sum of some terms of a geometric progression is $728$. If the common ratio is $3$ and the last term is $486$,what is the first term of the progression?

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