If the sum of first 20 terms of an A.P. is same as the sum of its first 28 terms, find the sum of its 48 terms.

Let first term of A.P. be a and common difference be d.

By formula,

Sum of first n terms of an A.P. = $\dfrac{n}{2}[2a + (n - 1)d]$

Given,

S₂₀ = S₂₈

$\therefore \dfrac{20}{2}[2a + (20 - 1)d] = \dfrac{28}{2}[2a + (28 - 1)d] \\[1em] \Rightarrow 10[2a + 19d] = 14[2a + 27d] \\[1em] \Rightarrow 20a + 190d = 28a + 378d \\[1em] \Rightarrow 28a - 20a = 190d - 378d \\[1em] \Rightarrow 8a = -188d \\[1em] \Rightarrow a = -\dfrac{188}{8}d \text{ ………(1)}$

Sum of 48 terms is

$S_{48} = \dfrac{48}{2}[2a + (48 - 1)d] \\[1em] = 24 \times [2 \times -\dfrac{188}{8}d + 47d] \\[1em] = 24 \times \Big[-\dfrac{376}{8}d + 47d\Big] \\[1em] = 24 \times \Big[-47d + 47d\Big] \\[1em] = 24 \times 0 \\[1em] = 0.$

Hence, S₄₈ = 0.