Answer : (b) 2
Sum of first 10 terms = \(\frac{1}{2}\) (Sum of next 10 terms)
⇒ S10 = \(\frac{1}{2}\) (S20 – S10) ⇒ 3S10 = S20
⇒ 3 × \(\frac{1}{2}\)[2a + 9d] = \(\frac{1}{2}\)[2a + 19d] \(\big[\because S_n = \frac{n}{2}[2a+(n-1)d]\big]\)
⇒ 3(2a + 9d) = 2(2a + 19d)
⇒ 6a + 27d = 4a + 38d
⇒ 2a = 11d
⇒ 2(13 + d) = 11d (∵ Second term = 13 and 13 – a = d)
⇒ 26 – 2d = 11d
⇒ 13d = 26
⇒ d = 2.