Question

In a certain A.P., the \[{{24}^{th}}\] term is twice the \[{{10}^{th}}\] term. Prove that the \[{{72}^{nd}}\] term is twice the \[{{34}^{th}}\] term.

Answer 1

Hint: Since the A.P is common, therefore the first term of an A.P. and the common difference of an AP will be the same. So we need to find a relation between \[{{a}_{1}}\] and \[d\].
Formula used:- \[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\] , where \[a\], is first term \[d\] is common difference.

Complete step by step solution:
As we know, the \[{{n}^{th}}\] term of an A.P. is given by the formula:- \[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\]
Where \[{{a}_{1}}\] is the first term and \[d\] is the common difference of the A.P.
Therefore, \[{{10}^{th}}\] term will be :- \[{{a}_{10}}={{a}_{1}}+\left( 10-1 \right)d\]
\[{{a}_{10}}={{a}_{1}}+9d\] -(i)
Thus, we get the \[{{10}^{th}}\] term of an A.P. as \[\left( {{a}_{1}}+9d \right)\]
and \[{{24}^{th}}\] term will be : \[{{a}_{24}}={{a}_{1}}+\left( 24-1 \right)d\]
\[{{a}_{24}}={{a}_{1}}+23d\] -(ii)
Thus, we get the \[{{24}^{th}}\] term of an AP as \[\left( {{a}_{1}}+23d \right)\]
Now, it is given that, the \[{{24}^{th}}\] term is twice the \[{{10}^{th}}\] term, thus we get;
\[\therefore {{a}_{24}}=2\times {{a}_{10}}\]
\[\therefore {{a}_{1}}+23d=2\times \left( {{a}_{1}}+9d \right)\] From (i) and (ii)
\[{{a}_{1}}+23d=2{{a}_{1}}+18d\]
\[\therefore 2{{a}_{1}}-{{a}_{1}}=23d-18d\]
\[\therefore {{a}_{1}}=5d\] -(iii)
Therefore \[{{1}^{st}}\] term of AP is \[5\] times the common difference.
Now, \[{{34}^{th}}\] term is:-
\[{{a}_{34}}={{a}_{1}}+\left( 34-1 \right)d\]
\[{{a}_{34}}={{a}_{1}}+33d\] -(iv)
\[\therefore \]Thus, we get \[{{34}^{th}}\] term, as \[\left( {{a}_{1}}+33d \right)\]
Put value of \[{{a}_{1}}\] in equation (iv)
\[{{a}_{34}}=5d+33d\] -(From iii)
\[\therefore {{a}_{34}}=38d\] -(v)
Also, \[{{34}^{th}}\]term of an AP is \[38\] times the common difference of AP.
Similarly, \[72nd\] term is:-
\[{{a}_{72}}={{a}_{1}}+\left( 72-1 \right)d\]
\[{{a}_{72}}={{a}_{1}}+\left( 71 \right)d\] -(vi)
Here, we get \[72nd\] term as \[\left( {{a}_{1}}+71d \right)\]
Write value of \[{{a}_{1}}\] in equation (vi)
\[{{a}_{72}}=5d+71d\] (From (iii)
\[\therefore {{a}_{72}}=76d\]
\[{{a}_{72}}=2\times 38d\]
Also, \[{{72}^{nd}}\] term is \[76\] times the common difference of AP.
From equation (v), we get
\[{{a}_{72}}=2\times {{a}_{34}}\] \[\left( \because 38d={{a}_{34}} \right)\]
\[\therefore {{72}^{nd}}\] term is twice the \[{{34}^{th}}\] term of given A.P.

Note: While solving such problems, we must be careful while defining \[{{a}_{n}}^{th}\] term. Also, we have to be careful while taking the first term of the A.P. as same throughout. To simplify, find the relation between \[{{a}_{1}}\] and \[d\], then put the value in required \[{{a}_{n}}\] and find the relation.