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8. Sequences and Series
easy
The sum of few terms of any ratio series is $728$, if common ratio is $3$ and last term is $486$, then first term of series will be
A
$2$
B
$1$
C
$3$
D
$4$
Solution
(a) $\therefore {n^{th}}$ term of series $ = a{r^{n – 1}}$$ = a\,{(3)^{n – 1}} = 486$ …..$(i)$
and sum of $n$ terms of series.
${S_n} = \frac{{a({3^n} – 1)}}{{3 – 1}}$ …..$(ii)$
From $(i),$ $a\left( {\frac{{{3^n}}}{3}} \right) = 486$ or $a{.3^n} = 3 \times 486 = 1458$
From $(ii),$ $a{.3^n} – a = 728 \times 2$ or $a{.3^n} – a = 1456$
$1458 – a = 1456$
$⇒$ $a = 2$.
Standard 11
Mathematics
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