In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

Given: 24th term is twice the 10th term

a24 = 2a10

To prove: a72 = 2a34

We know, an = a + (n – 1)d where a is first term or a1 and d is common difference and n is any natural number

When n = 10:

a10 = a + (10 – 1)d

a10 = a + 9d

When n = 24:

a24 = a + (24 – 1)d

a24 = a + 23d

When n = 34:

a34 = a + (34 – 1)d

a34 = a + 33d ………(i)

When n = 72:

a72 = a + (72 – 1)d

a72 = a + 71d

According to question:

a24 = 2a10

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

a + 23d = 2a + 18d

a – 2a + 23d – 18d = 0

-a + 5d = 0

a = 5d

Now, a72 = a + 71d

a72 = 5d + 71d

a72 = 76d

a72 = 10d + 66d

a72 = 2(5d + 33d)

{ a = 5d}

a72 = 2(a + 33d)

a72 = 2a34 (From (i))

Hence Proved

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