In a certain A.P. the 24th term is twice the 10th term.

Here, we are given that 24th term is twice the 10th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d

We have to prove that 

So, let us first find the two terms.

As we know,

$a_{n}=a+(n-1) d$

For $10^{\text {th }}$ term $(n=10)$,

$a_{10}=a+(10-1) d$

$=a+9 d$

For $24^{\text {th }}$ term $(n=24)$,

$a_{24}=a+(24-1) d$

$=a+23 d$

Now, we are given that $a_{24}=2 a_{10}$

So, we get,

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

$a+23 d=2 a+18 d$

$23 d-18 d=2 a-a$

$5 d=a$ .......(1)

Further, we need to prove that the 72nd term is twice of 34th term. So let now find these two terms,

For 34th term (n = 34),

$a_{34}=a+(34-1) d$

$=5 d+33 d \quad$ (Using 1)

$=38 d$

For 72nd term (n = 72),

$a_{72}=a+(72-1) d$

$=5 d+71 d$

$=76 d$ (Using 1)

$=2(38 d)$

Therefore, $a_{72}=2 a_{34}$