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In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
Given:
In a certain A.P. the 24th term is twice the 10th term.
To do:
We have to prove that the 72nd term is twice the 34th term.
Solution:
Let the required A.P. be $a, a+d, a+2d, ......$
Here,
$a_1=a, a_2=a+d$ and Common difference $=a_2-a_1=a+d-a=d$
We know that,
$a_n=a+(n-1)d$
Therefore,
$a_{24}=a+(24-1)d$
$=a+23d$.....(i)
$a_{10}=a+(10-1)d$
$=a+9d$....(ii)
According to the question,
$a_{24}=2\times a_{10}$
$a+23d=2(a+9d)$ (From(i) and (ii))
$a+23d=2a+18d$
$2a-a=23d-18d$
$a=5d$.....(iii)
This implies,
34th term $a_{34}=a+(34-1)d$
$=a+33d$
$=5d+33d$ (From (iii))
$=38d$.....(iv)
72nd term $a_{72}=a+(72-1)d$
$=a+71d$
$=5d+71d$ (From (iii))
$=76d$
$=2(38d)$
$=2\times a_{34}$ (From (iv))
Therefore, 72nd term is twice the 34th term.
Hence proved.
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