# Find the sum of each of the following APs:

Question:

Find the sum of each of the following APs:

(i) $2,7,12,17, \ldots$ to 19 terms

(ii) $9,7,5,3, \ldots$ to 14 terms

(iii) $-37,-33,-29, \ldots$ to 12 terms

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

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

Solution:

(i) The given AP is 2, 7, 12, 17, ... .

Here, a = 2 and d = 7 − 2 = 5

Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we have

$S_{19}=\frac{19}{2}[2 \times 2+(19-1) \times 5]$

$=\frac{19}{2} \times(4+90)$

$=\frac{19}{2} \times 94$

$=893$

(ii) The given AP is 9, 7, 5, 3, ... .

Here, a = 9 and d = 7 − 9 = −2

Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we have

$S_{14}=\frac{14}{2}[2 \times 9+(14-1) \times(-2)]$

$=7 \times(18-26)$

$=7 \times(-8)$

$=-56$

(iii) The given AP is −37, −33, −29, ... .

Here, a = −37 and d = −33 − (−37) = −33 + 37 = 4

Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we have

$S_{12}=\frac{12}{2}[2 \times(-37)+(12-1) \times 4]$

$=6 \times(-74+44)$

$=6 \times(-30)$

$=-180$

(iv) The given AP is $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$.

Here, $a=\frac{1}{15}$ and $d=\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}$

Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we have

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

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

$=\frac{11}{2} \times\left(\frac{18}{60}\right)$

$=\frac{33}{20}$

(v) The given AP is 0.6, 1.7, 2.8, ... .

Here, a = 0.6 and d = 1.7 − 0.6 = 1.1

Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we have

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

$=50 \times(1.2+108.9)$

$=50 \times 110.1$

$=5505$