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.

To do:

We have to find the sum of the given APs.

Solution:

(i) $a=2, d=7-2=5, n=10$

We know that,

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

$S_{10}=\frac{10}{2}[2 \times 2+(10-1) 5]$

$=5(4+45)$

$=5 \times 49$

$=245$

(ii) $a=-37, d=-33-(37)=-33+37=4, n=12$

We know that,

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

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

$=6(-74+44)$

$=6 \times(-30)$

$=-180$

(iii) $a=0.6, d=1.7-0.6=1.1, n=100$

We know that,

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

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

$=50(1.2+108.9)$

$=50 \times 110.1$

$=5505$

(iv) $a=\frac{1}{15}, d=\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}, n=11$

We know that,

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

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

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

$=\frac{11}{2}(\frac{8+10}{60})$

$=\frac{11}{2} \times \frac{18}{60}$

$=\frac{33}{20}$

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Updated on: 10-Oct-2022

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