# Find the sum of the following APs :

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) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ to 11 terms

Solution:

(i) $a=2, d=5$

$S_{10}=\frac{10}{2}\{2 a+9 d\}$

$\left(\because \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\{2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}\}\right)$

$=5 \times\{2 \times 2+9 \times 5)=5 \times 49=245$

(ii) $\mathrm{a}=-37, \mathrm{~d}=4$

$S_{12}=\frac{12}{2}\{2 a+11 d\}$

$=6 \times\{2(-37)+11 \times 4\}$

$=6 \times\{-74+44\}=-180$

(iii) $0.6,1.7,2.8, \ldots .$, to 100 terms

For this A.P.,

$a=0.6$

$d=a_{2}-a_{1}=1.7-0.6=1.1$

$n=100$

$S_{100}=\frac{100}{2}[2(0.6)+(100-1) 1.1]$

= 50[1.2 + (99) × (1.1)]

= 50[110.1]

= 110.1

= 5505

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

For this A.P.,

$a=\frac{1}{15}$

$\mathrm{n}=11$

$d=a_{2}-a_{1}=\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}$

We know that

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

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

$=\frac{11}{2}\left[\frac{2}{15}+\frac{10}{60}\right]$

$=\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right]=\frac{11}{2}\left[\frac{4+5}{30}\right]$

$=\left(\frac{11}{2}\right)\left(\frac{9}{30}\right)=\frac{33}{20}$