Question

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

Answer 1

Given:

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

To prove:

72nd term is 4 times its 15th term

Solution:

Consider a to be the 1st  term and d to be the common difference of the AP.

From given,

$a_{24}=2 a_{10} \rightarrow(i)$

We know that, $a_{n}=a+(n-1) d \rightarrow(i i)$

Thus, $a_{24}=a+23 d \text { and } a_{10}=a+9 d$

Substitue the value of $a_{24}\ and a_{10}$ in equation (i), we get

$\begin{array}{l}{a+23 d=2(a+9 d)} \\ {a+23 d=2 a+18 d}\end{array}$

Now, shift the a terms to right hand side and keep the d terms in left hand side, we get

$\begin{array}{c}{23 d-18 d=2 a-a} \\ {\Rightarrow a=5 d}\end{array}$

Now, from the question,

$\frac{a_{72}}{a_{15}}=\frac{a+71 d}{a+14 d}$

Now, substitute the value of a=5 in above equation, we get

$\begin{array}{c}{\frac{a_{72}}{a_{15}}=\frac{76 d}{19 d}}\\ \\ {\frac{a_{72}}{a_{15}}=4} \\\\ {\Rightarrow a_{72}=4\left(a_{15}\right)}\end{array}$

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

Hence proved.

Answer 2

Answer:

Let a , d are first term and common difference of an A.P.

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

/* According to the problem given,

$24^{th}\:term = 2\times 10^{th}\:term$

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

$\implies a+23d = 2(a+9d)$

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

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

$\implies 5d = a \:--(1)$

$Now, \\72{nd} \:term \\= a+71d\\=5d+71d$

/* from (1) */

$=76d\\=4\times 19d\\=4\times (5d+14d)\\=4(a+14d)$

/* Form (1)*/

$=4a_{15}$

Therefore,.

$72^{nd}\:term = 4\times 15^{th}\:term$

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