If S_n = 4 + 11 + 21 + 34 + 50 + to n terms, then 160( S_29 - S_9 ) is equal to

(a) : Given, S_n=4+11+21+34+ . up to n terms \, S_n=4+11+21++ up to n terms \ (ii) from (i), we get 0=[4+7+10+13++]-a_n \\ a_n=n2[2(4)+(n-1)(3)] \\=n2[3 n+5]=3 n^2+5 n2\\ Now, S_n= a_n=

If S_{n} = 4 + 11 + 21 + 34 + 50 + to n terms, then \frac{1}{60}( S_{29} - S_{9} ) is equal to: Sequence & Series (Mathematics)

(a) : Given, $S_n=4+11+21+34+\cdots \cdots$ . up to $n$ terms $\\$ Also, $S_n=4+11+21++$ up to $n$ terms $\\$ Subtracting (ii) from (i), we get $0=[4+7+10+13++]-a_n \\ \therefore a_n=\frac{n}{2}[2(4)+(n-1)(3)] \\=\frac{n}{2}[3 n+5]=\frac{3 n^2+5 n}{2}\\$ Now, $S_n=\sum a_n=\frac{3}{2} \sum n^2+\frac{5}{2} \sum n$ $\\ =\frac{3}{2} \frac{n \cdot(n+1)(2 n+1)}{6}+\frac{5}{2} \frac{n(n+1)}{2} \\=\frac{n(n+1)}{4}[(2 n+1)+5] \\ =\frac{n(n+1)}{4} \frac{(2 n+6)}{2}=\frac{n(n+1)(n+3)}{2}\\$ Now, $\frac{1}{60}\left[S_{29}-S_9\right] \\=\frac{1}{60}\left[\frac{(29)(30)(32)}{2}-\frac{(9)(10)(12)}{2}\right]$ $\\=\frac{1}{60}[13920-540]=\frac{13380}{60}=223$

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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