Question

The sum of the first six terms of an Ap is 42.the ratio of its 10th term to its 30th term is 1:3.find the first term of an Ap

Answer 1

AnSwer:-

• Let a be the first term and d be the common difference.

$\longmapsto \sf \: S _{6} = 42$

$\sf \longmapsto \: \dfrac{6}{2}(2a + 5d) = 42 \\ \\ \sf \longmapsto \: 2a + 5d = 14 \\ \\ \sf \longmapsto \: 2a = 14 - 5d \\ \\ \sf \longmapsto \: \dfrac{a _{10} }{a _{30}} = \dfrac{1}{3} \\ \\ \sf \longmapsto \: \dfrac{a + 9d}{a + 29d} = \dfrac{1}{3} \\ \\ \sf \longmapsto \: \: \dfrac{2a + 18d}{2a + 58d} = \dfrac{1}{3} \\ \\ \sf \longmapsto \: \dfrac{14 - 5d + 18d}{15 - 5d + 58d} = \dfrac{1}{3} ...........(from \: 1) \\ \\ \sf \longmapsto \: \dfrac{14 + 13d}{14 + 53d} = \dfrac{1}{3} \\ \\ \sf \longmapsto \:42 + 39d = 14 + 32d \\ \\ \sf \longmapsto \:28= 14d \\ \\ \sf \longmapsto \:d = 2 \\ \\ \sf \: from \: (1) \\ \\ \sf \longmapsto \:2a + 5d = 14 \\ \\ \sf \longmapsto \:2a + 5(2) = 14 \\ \\ \sf \longmapsto \:2a + 10 = 14 \\ \\ \sf \longmapsto \:2a = 14 - 10 \\ \\ \sf \longmapsto \:2a = 4 \\ \\ \sf \longmapsto \:a = 2$

Now,

$\sf \longmapsto \:a _{13} = 2 + 12 \times 2 \\ \\ \sf \longmapsto \:a _{13} = 14 \times 2 \\ \\ \sf \longmapsto \:a _{13} = 26$