The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is 4 times its 15th term.

Given:

The 24th term of an A.P. is twice its 10th term. 

To do:

We have to show that the 72nd term is 4 times the 15th 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,

15th term $a_{15}=a+(15-1)d$

$=a+14d$

$=5d+14d$    (From (iii))

$=19d$.....(iv)

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

$=a+71d$

$=5d+71d$    (From (iii))

$=76d$

$=4(19d)$

$=4\times a_{15}$     (From (iv))

Therefore, 72nd term is 4 times the 15th term.

Hence proved. 

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Updated on: 10-Oct-2022

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