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

In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th 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, $a_{72}=2 a_{34}$


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