Question

Find the sum of the following series:
(i) 2 + 5 + 8 + .... + 182
(ii) 101 + 99 + 97 + .... + 47
(iii) $(a-b{)}^2+(a^2+b^2)+(a+b{)}^2+\dots \dots +\left[\right(a+b{)}^2+6ab]$

Answer 1

(i) 2 + 5 + 8 + .... + 182
$a_n$ term of given A.P. is 182
$a_n=a+(n-1)d=182\Rightarrow 182=2+(n-1)3\Rightarrow 182=3n-1\Rightarrow 3n=183\Rightarrow n=61$
Then,
$S_n=\frac{n}{2}[a+l]=\frac{61}{2}[2+182]=61\times 92=5612$

(ii) 101 + 99 + 97 + .... + 47
$a_n$ term of A.P. of n terms is 47.
$\therefore 47=a+(n-1)d47=101+(n-1)(-2)\Rightarrow 101-2n+2=47\Rightarrow 2n-2=54$
Or n = 28
Then,
$S_n=\frac{n}{2}[a+l]=\frac{28}{2}[101+47]=14\times 148=2072$

(iii) $(a-b{)}^2+(a^2+b^2)+(a-b{)}^2+\dots \dots +\left[\right(a+b{)}^2+6ab]$
Here, the series is an A.P. where we have the following:
$a=(a-b{)}^{2}d=(a^2+b^2-(a-b{)}^2)=2ab{a}_n=\left[\right(a+b{)}^2+6ab]\Rightarrow (a-b{)}^2+(n-1)\left(2ab\right)=\left[\right(a+b{)}^2+6ab]\Rightarrow a^2+b^2-2ab+2abn-2ab=[a^2+b^2+2ab+6ab]\Rightarrow a^2+b^2-4ab+2abn=a^2+b^2+8ab\Rightarrow 2abn=12ab\Rightarrow n=6S_n=\frac{n}{2}[2a+(n-1)d]\Rightarrow S_6=\frac{6}{2}[2(a-b{)}^2+(6-1\left)2ab\right]=3\left[2\right(a^2+b^2-2ab)+10ab]=3[2a^2+2b^2-4ab+10ab]=3[2a^2+2b^2+6ab]=6[a^2+b^2+3ab]$