An A.P. has a property that the sum of the first ten terms is half the

1.	An A.P. has a property that the sum of the first ten terms is half the sum of the next ten terms. If the second term is 13, then the common difference is(a) 3 (b) 2 (c) 5 (d) 4
Answer» 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.

An A.P. has a property that the sum of the first ten terms is half the sum of the next ten terms. If the second term is 13, then the common difference is(a) 3 (b) 2 (c) 5 (d) 4

Answer»

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.