If the sum of the first $n$ terms of an A.P is given $Sn$= $2n^2+5n$. Find the $n$th term of the A.P ?
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use the fact that $S_{n-1}+T_n=S_{n}$ – Rishi Dec 01 '16 at 02:42
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$a_n = nd+a_0$
$S_n = $$d\sum_\limits{i=1}^n k + na_1\\ d(\frac12)(n^2 + n) + na_0\\\frac d2 = 2, \frac d2 + a_0 = 5\\ d = 4, a_0 = 3\\ a_n = 4n + 3$
Alternate:
$S_n = S_{n-1} + a_n\\ 2n^2 + 5n = 2(n-1)^2 + 5(n-1) + a_n\\ 2n^2 + 5n = 2n^2 - 4n + 2 + 5n - 5 + a_n\\ 4n+3 = a_n$
Doug M
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How did you know $Sn=Sn-1+an$. How you worte this line please explain, because i couldn't understand it. – changer Dec 01 '16 at 14:55
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$S_n = \sum_\limits{i=1}^n a_i = (\sum_\limits{i=1}^{n-1} a_i) + a_n = S_{n-1} + a_n$ – Doug M Dec 01 '16 at 16:27