Question

Find the sum of the following APs:
(i) $2,7,12,...$, to $10$ terms.
(ii) $37,33,29,...$, to $12$ terms.
(iii) $0.6,1.7,2.8,...$, to $100$ terms.
(iv) $\frac{1}{15},\frac{1}{12},\frac{1}{10}$,...., to 11 terms.

Answer 1

(i) $2,7,\mathrm{12......}$ to ${10}^{th}$ term

Here $a=2,n=10$

And $d=7-2=5$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_{10}=\frac{10}{2}\left[2\right(2)+(10-1\left)5\right]=5(4+9\times 5)=245$

(ii) $37,33,29,........$ to ${12}^{th}$ term

Here $a=37,n=12$

And $d=33-37=-4$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_{12}=\frac{12}{2}\left[2\right(37)+(12-1\left)\right(-4]=6(74-11\times 4)=-180$

(iii) $0.6,1.7,\mathrm{2.8............}$ to $100$ term

Here $a=0.6,n=100$

And $d=1.7-0.6=1.1$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_{100}=\frac{100}{2}\left[2\right(0.6)+(100-1\left)\right(1.1\left)\right]=50(1.2-99\times 1.1)=5505$

(iv)$\frac{1}{15},\frac{1}{12},\frac{1}{10}.......$ to ${11}^{th}$ term

$a=\frac{1}{15}$, $n=11$

$d=\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}$

$S_{11}=\frac{11}{2}\left[2\right(\frac{1}{15}+(11-1)\frac{1}{60}\left)\right]$

$=$ $\frac{11}{2}(\frac{2}{15}+\frac{10}{60})$

$=$ $\frac{11}{2}\times \frac{9}{30}=\frac{33}{20}$