Find the sum of the following A.P.s :

(i) 2, 7, 12 … to 10 terms

(ii) $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}, …$ to 11 terms

(i) Here, a = 2, d = 7 - 2 = 5 and n = 10.

We know that Sum = $\dfrac{n}{2}[2a + (n - 1)d]$

$\therefore \text{ Sum } = \dfrac{10}{2}[2 \times 2 + (10 - 1)5] \\[1em] = 5[4 + 9 \times 5] \\[1em] = 5 \times 49 \\[1em] = 245.$

Hence, the sum of the A.P. 2, 7, 12, … upto 10 terms is 245.

(ii) Here, a = $\dfrac{1}{15}, d = \dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5 - 4}{60} = \dfrac{1}{60}$ and n = 11.

We know that Sum = $\dfrac{n}{2}[2a + (n - 1)d]$

$\therefore \text{ Sum } = \dfrac{11}{2}[2 \times \dfrac{1}{15} + (11 - 1)\dfrac{1}{60}] \\[1em] = \dfrac{11}{2}\Big[\dfrac{2}{15} + \dfrac{10}{60}\Big] \\[1em] = \dfrac{11}{2} \times \Big[\dfrac{2 \times 4 + 10}{60}\Big] \\[1em] = \dfrac{11}{2} \times \dfrac{18}{60}$

Dividing numerator and denominator by 6,

$= \dfrac{198}{120} \\[1em] = \dfrac{33}{20}. \\[1em]$

Hence, the sum of the A.P. $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}, …$ upto 11 terms is $\dfrac{33}{20}$.