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Arithmetic Mean

In an arithme...

Question

In an arithmetic progression ${a_{n}}, a_{1} > 0$ and $3 a_{8} = 5 a_{13}$ .
Let $S_{n}$ be the sum to first ‘n’ terms. The value of ‘n’ for which $S_{n}$ is maximum is

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Solution

The correct option is C 20
Let ‘d’ be the common difference $2 a_{1} + 39 d = 0$
Sn is maximum $\Rightarrow a_{x + 1} < 0$
$a_{1} + n \frac{(- 2 a_{1})}{39} = (39 - 2 x) \frac{a_{1}}{39} < 0$
$n > 19 \Rightarrow n + 1 > 20 \Rightarrow n + 1 \geq 21$
$∴ n = 20$

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In an arithmetic progression {an},a1>0 and 3a8=5a13. Let Sn be the sum to first ‘n’ terms. The value of ‘n’ for which Sn is maximum is

The correct option is C 20Let ‘d’ be the common difference 2a1+39d=0 Sn is maximum ⇒ ax+1<0 a1+n(−2a1)39=(39–2x)a139<0 n>19⇒ n+1>20⇒ n+1≥21 ∴ n=20

In an arithmetic progression ${a_{n}}, a_{1} > 0$ and $3 a_{8} = 5 a_{13}$ .
Let $S_{n}$ be the sum to first ‘n’ terms. The value of ‘n’ for which $S_{n}$ is maximum is

The correct option is C 20
Let ‘d’ be the common difference $2 a_{1} + 39 d = 0$
Sn is maximum $\Rightarrow a_{x + 1} < 0$
$a_{1} + n \frac{(- 2 a_{1})}{39} = (39 - 2 x) \frac{a_{1}}{39} < 0$
$n > 19 \Rightarrow n + 1 > 20 \Rightarrow n + 1 \geq 21$
$∴ n = 20$