Question

If (m + 1)^th term of an A.P. is twice the (n + 1)^th term, then prove that (3m + 1)^th term is twice the (m + n + 1)^th term

Answer 1

t_n = a + (n – 1)d

t_m+1 = a + (m + 1 – 1)d

= a + md

t_n+1 = a + (n + 1 – 1)d

= a + nd

2(t_n+1) = 2(a + nd)

t_m+1 = 2t_n+1    ...(1)

⇒ a + md = 2(a + nd)

2a + 2nd – a – md = 0

a + (2n – m)d = 0

t(_3m+1) = a + (3m + 1 – 1)d

= a + 3md

t(_m+n+1) = a + (m + n + 1 – 1)d

= a + (m + n)d

`2("t"_("m"+ "n" + 1))` = 2(a + (m + n)d)

= 2a + 2md + 2nd

`"t"_((3"m" + 1)) = 2"t"_(("m" + "n" + 1))`    ...(2)

a + 3md = 2a + 2md + 2nd

2a + 2md + 2nd – a – 3md = 0

a – md + 2nd = 0

a + (2n – m)d = 0

∴ It is proved that `"t"_((3"m" + 1)) = 2"t"_(("m" + "n" + 1))`