Question

In an AP 24th term is twice 10th term.Show that 72nd term is twice 34th term

Answer 1

According to question,

T24 = 2 × T10

=> a + 23d = 2 ( a + 9d)

=> a + 23d = 2a + 18d

=> 23d - 18d = 2a - a

=> a = 5d

Now,

T72 = a + 71d

=> T72 = 5d + 71d

=> T72 = 76d --------(1)

Now,

T34 = a + 33d

=> T34 = 5d + 33d

=> T34 = 38d ------(2)

On Comparing equation 1 and 2, we get

T72 = 2 × T34


Hence Proved......

Answer 2

Solution:-
let first term is (a) and common diffrence is (d)
acording to quetions:-


$t(24) = 2 \times t(10) \\ a + (n - 1)d = 2 \times(a +( n - 1)d) \\ a + 23d = 2a + 18d \\ a = 5d...........(1) \\ now \\ \\t(34) = a + (n - 1)d \\ by \: eq(1) \\ t(34) = 5d + 33d \\ t(34) = 38d \\ now \\ t(72) = t(72) = a +( n - 1)d \\ by \: eq(1) \\ t(72) = 5d + 71d \\ t(72) = 76d \\ t(72) = 2 \times 38d \\ t(72) = 2 \times t(34) \\ proved$