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


The $24^{\text {th }}$ term of an A.P. is twice its $10^{\text {th }}$ term. Show that its $72^{\text {nd }}$ term is 4 times its $15^{\text {th }}$ term.


Let a be the first term and d be the common difference.

We know that, $n^{\text {th }}$ term $=a_{n}=a+(n-1) d$

According to the question,

$a_{24}=2 a_{10}$

$\Rightarrow a+(24-1) d=2(a+(10-1) d)$

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


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

$\Rightarrow 5 d=a$


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


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

    = 5d + 71d              [From (1)]
      = 76d                      ..... (2)


a15 a + (15 − 1)d
      = 5d + 14d              [From (1)]
      = 19d                      ..... (3)

On comparing (2) and (3), we get

$76 d=4 \times 19 d$

$\Rightarrow a_{72}=4 \times a_{15}$

Thus, $72^{\text {nd }}$ term of the given A.P. is 4 times its $15^{\text {th }}$ term.



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