Question

the 24th term of an AP is twice is 10th term. Show that its 72nd term is 4 times its 15 the term.

Answer 1

Let first term = a

and common difference = d

we know that $a_n=a+(n-1)d$

Given that, $a_{24}=2\times a_{10}$

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

$\Rightarrow a+23d=2\times (a+9d)$

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

$\Rightarrow a=5d....\left(i\right)$

Now 15th term, $a_{15}=a+(15-1)d=a+14d\left[\text{From (i)}\right]=5d+14d=19d$

Similarly 72nd term, $a_{72}=a+(72-1)d=a+71d=5d+71d=76d$

Now,
$\frac{a_{72}}{a_{15}}=\frac{76d}{19d}=4$

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

Hence, ${72}^{th}$ term is $4$ times to its ${15}^{th}$ term.