Find the sum :

$\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}$, ……to 11 terms.

We know that,

Sum of n terms of an A.P. is given by,

∴ S_n = $\dfrac{n}{2}$ [2a + (n - 1)d]

Given,

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

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

n = 11

$\Rightarrow S_{11} = \dfrac{11}{2} \Big[2\Big(\dfrac{1}{15}\Big) + (11 - 1)\Big(\dfrac{1}{60}\Big)\Big] \\[1em] = \dfrac{11}{2} \Big[\Big(\dfrac{2}{15}\Big) + \Big(\dfrac{10}{60}\Big)\Big] \\[1em] = \dfrac{11}{2} \Big[\dfrac{2 \times 2 + 1 \times 5 }{30}\Big] \\[1em] = \dfrac{11}{2} \Big[\dfrac{4 + 5}{30}\Big] \\[1em] = \dfrac{11}{2} \Big[\dfrac{9}{30}\Big] \\[1em] = \dfrac{11}{2} \Big[\dfrac{3}{10}\Big] \\[1em] = \dfrac{33}{20}.$

Hence, S₁₁ = $\dfrac{33}{20}$.