Let a1- a2- a3- - a100 be an arithmetic progression with a1 - 3 and Sp-i-1pai- 1- p - 100 - For any integer n with 1 - n- 20- let m - 5n- If SmSn does not depend on n- then a2 is

We have, a_1 = 3, S_p = a_1 + a_2 + a_3 + ...... + a_p, 1 ≤ p≤ 100 By the given condition, S_mS_n=mn6+(m 1)d6+(n 1)d=k , which is independent of m and n ⇒ nbs ...

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Standard XII

Mathematics

Common Roots

Let a1, a2, a...

Question

Let $a_{1}, a_{2}, a_{3}, . . ., a_{100}$ be an arithmetic progression with $a_{1} = 3$ and $S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100$ . For any integer $n$ with $1 \leq n \leq 20$ , let $m = 5 n$ . If $\frac{S_{m}}{S_{n}}$ does not depend on $n$ , then $a_{2}$ is

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Solution

We have,
$a_{1} = 3, S_{p} = a_{1} + a_{2} + a_{3} + . . . . . . + a_{p}, 1 \leq p \leq 100$
By the given condition,
$\frac{S_{m}}{S_{n}} = \frac{m}{n} \frac{6 + (m - 1) d}{6 + (n - 1) d} = k$ , which is independent of $m$ and $n$
$\Rightarrow 5 (6 + (m - 1) d) = k (6 + (n - 1) d)$
$\Rightarrow 30 + 25 n d - 5 d = 6 k + n k d - k d$
which is independent of $n$
Hence, on comparing the coefficients of like terms, we get
$k = 25$
$30 - 5 d = 150 - 25 d$
$\Rightarrow 20 d = 120$
$\Rightarrow d = 6$
$a_{2} = 9$

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Let a1,a2,a3,...,a100 be an arithmetic progression with a1=3 and Sp=∑pi=1ai,1≤p≤100 . For any integer n with 1≤n≤20, let m=5n. If SmSn does not depend on n, then a2 is

Let $a_{1}, a_{2}, a_{3}, . . ., a_{100}$ be an arithmetic progression with $a_{1} = 3$ and $S_{p} = \sum_{i = 1}^{p} a_{i}, 1 \leq p \leq 100$ . For any integer $n$ with $1 \leq n \leq 20$ , let $m = 5 n$ . If $\frac{S_{m}}{S_{n}}$ does not depend on $n$ , then $a_{2}$ is