Question

Find the sum of first n terms of an A.P. whose nth term is 5n-1. Hence find the sum of first 20 terms ?

Answer 1

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Answer 2

Answer:

$S_{n}=\frac{5n^{2}+3n}{2}\\S_{20}=1030$

Step-by-step explanation:

$Given\\n^{th}\:term \:of \:A.P = a_{n}=5n-1$

$Sum\:of \: first \:n\:terms \\=S_{n}=\sum a_{n}\\=\sum (5n-1)\\=5\sum n - \sum 1\\=5\left(\frac{n(n+1)}{2}\right)-n$

$Since, \sum n = \frac{n(n+1)}{2}$

$=\frac{5n^{2}+5n}{2}-n$

$=\frac{5n^{2}+5n-2n}{2}$

$S_{n}=\frac{5n^{2}+3n}{2}\:---(1)$

$ii) Sum\:of \:first\:20\:terms\\ = S_{20}$

/* Substitute n=20 in equation (1), we get

$S_{20}=\frac{5\times (20)^{2}+3\times 20}{2}\\=\frac{2000+60}{2}\\=\frac{2060}{2}\\=1030$

Therefore,

$S_{n}=\frac{5n^{2}+3n}{2}\\S_{20}=1030$

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