Let a_1, a_2, a_3, , a 100 be an arithmetic progression with a_1=3 and S_p=_i=1^p a_i, 1 p 100. For any integer n with 1 n 20, t m=5 n. If S_mS_n does not depend on .n then a_2 is \_\_\_\_ .

s_ms_n=s_5 ns_n=5 n2(6+(5 n-1) d)n2(6+(n-1) d) \\ = 5((6-d)+5 n d)(6-d)+n d \\ For d=6 we have S_5 n/S_n independent of ' n '. \, For d=0, it's trivially true. So, a_2=9&nbsp

Let a_{1},a_{2},a_{3},.,a_{n} be an arithmetic progression with a_{1} = 3 and S_{p} = \sum_{i = 1}^{p}\mspace{2mu} a_{i},1 \leq p \leq 100. For any integer n with 1 \leq n \leq 20. Let m = 5n. If \frac{S_{m}}{S_{n}} does not depend on n, then a_{2} is n, then a_{2} is: Sequence & Series (Mathematics)

$\frac{s_m}{s_n}=\frac{s_{5 n}}{s_n}=\frac{\frac{5 n}{2}(6+(5 n-1) d)}{\frac{n}{2}(6+(n-1) d)} \\ = \frac{5((6-d)+5 n d)}{(6-d)+n d}$ $\\$ For $d=6$ we have $S_{5 n}/S_n$ independent of ' $n$ '. $\\$ Also, For $d=0$ , it's trivially true. So, $a_2=9$ $\\$ Note : This question has two possible answers. $\\$ So although the question is correct, answer is not unique

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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