Question

If ${{\left( m+1 \right)}^{th}}$ term of an A.P. is twice the ${{\left( n+1 \right)}^{th}}$ term, prove that the ${{72}^{nd}}$ term is twice the ${{34}^{th}}$ term?

Answer 1

We know that the general term of an A.P. is as follows: ${{T}_{n}}=a+\left( n-1 \right)d$. Then put n as $\left( m+1 \right)$ in this general term and get the ${{\left( m+1 \right)}^{th}}$ term. Similarly, substitute n as $\left( n+1 \right)$ in the general term and get the ${{\left( n+1 \right)}^{th}}$ and then use the relation that ${{\left( m+1 \right)}^{th}}$ term of an A.P. is twice the ${{\left( n+1 \right)}^{th}}$. From this, you will get the relation between $''a\And d''$. Then we have to prove ${{72}^{nd}}$ term is twice the ${{34}^{th}}$ term which we are going to do by substituting n as 72 in the general term of an A.P. and substitute n as 34 in the general term.

Complete step by step answer:
We know the general term of an A.P. is as follows:
${{T}_{n}}=a+\left( n-1 \right)d$
Writing the ${{\left( m+1 \right)}^{th}}$ term by substituting n as $\left( m+1 \right)$ in the above general term and we get,
$\begin{aligned} & {{T}_{m+1}}=a+\left( m+1-1 \right)d \\\ & \Rightarrow {{T}_{m+1}}=a+md \\\ \end{aligned}$
Writing the ${{\left( n+1 \right)}^{th}}$ term by substituting n as $\left( n+1 \right)$ in the above general term and we get,
$\begin{aligned} & {{T}_{n+1}}=a+\left( n+1-1 \right)d \\\ & \Rightarrow {{T}_{n+1}}=a+nd \\\ \end{aligned}$
It is given that ${{\left( m+1 \right)}^{th}}$ term of an A.P. is twice the ${{\left( n+1 \right)}^{th}}$ term so using this relation in the above and we get,
${{T}_{m+1}}=2{{T}_{n+1}}$
Substituting the values of ${{T}_{m+1}}\And {{T}_{n+1}}$ in the above and we get,
$\begin{aligned} & a+md=2\left( a+nd \right) \\\ & \Rightarrow a+md=2a+2nd \\\ & \Rightarrow a=md-2nd \\\ \end{aligned}$
Taking “d” as common the R.H.S of the above equation and we get,
$a=\left( m-2n \right)d$ …………… (1)
Now, we are going to find ${{72}^{nd}}$ term and ${{34}^{th}}$ term by substituting n as 72 and 34 respectively in the general term of an A.P.
$\begin{aligned} & {{T}_{n}}=a+\left( n-1 \right)d \\\ & \Rightarrow {{T}_{72}}=a+\left( 72-1 \right)d \\\ & \Rightarrow {{T}_{72}}=a+71d \\\ \end{aligned}$
Calculating ${{34}^{th}}$ term we get,
$\begin{aligned} & {{T}_{n}}=a+\left( n-1 \right)d \\\ & \Rightarrow {{T}_{34}}=a+\left( 34-1 \right)d \\\ & \Rightarrow {{T}_{34}}=a+33d \\\ \end{aligned}$
We have to show that the ${{72}^{nd}}$ term is twice the ${{34}^{th}}$ term so writing these two terms as follows:
$\begin{aligned} & {{T}_{72}}=2{{T}_{34}} \\\ & \Rightarrow a+71d=2\left( a+33d \right) \\\ & \Rightarrow a+71d=2a+66d \\\ \end{aligned}$
Rearranging the above equation we get,
$a=71d-66d$
We can write the above equation as follows:
$a=\left( 71-2\left( 33 \right) \right)d$ ………….. (2)
From eq. (1) we got the following:
$a=\left( m-2n \right)d$
This is true when the two terms are ${{\left( m+1 \right)}^{th}}$ and ${{\left( n+1 \right)}^{th}}$ so using this relation we have found the general term for 72nd term and 34th so it can be written as ${{\left( 71+1 \right)}^{th}}\And {{\left( 33+1 \right)}^{th}}$ respectively then m and n will become 71 and 33 in the above equation (1) and we get,
$a=\left( 71-2\left( 33 \right) \right)d$
This is the same relation which we are getting in eq. (1) so we have proved that the ${{72}^{nd}}$ term is twice the ${{34}^{th}}$ term.

Note: It will be impossible for you to move forward in this problem if you don’t know the general term of an A.P. so make sure you correctly know the formula for the general term of an A.P. Also, the mistake that could be possible in the above problem is that you might miss $-1$ in the general term of an A.P.
And the miss makes the general term of an A.P. will look like:
${{T}_{n}}=a+\left( n \right)d$
So, make sure you will write the correct formula.