Question

Find the $11^{th}$ terms from the last term of the A.P. as 27, 23, 19, 25, ...., -65.

Answer 1

First, we should know the formula to calculate the $n^{th}$ term of the A.P. as ${{a}_{n}}=a+\left( n-1 \right)d$. Then, we need to find the $11^{th}$ term from the end, so we reverse the A.P as -65, -61, -57, ....., 27.Then, by substituting the value of a as -65 and value of d as 4 calculated to get the value of ${{a}_{11}}$.

Complete step-by-step answer:
In this question, we are supposed to the $11^{th}$ terms from the last term of the A.P. as 27, 23, 19, ...., -65.
So, we should know the formula to calculate the $n^{th}$ term of the A.P. as:
${{a}_{n}}=a+\left( n-1 \right)d$
Here, in the above formula ${{a}_{n}}$ is the $n^{th}$ term of A.P where a is first term and d is the difference between the two consecutive terms.
Now, we need to find the $11^{th}$ term from the end, so we reverse the A.P as -65, -61, -57, ....., 27.
So, a for the above given series is -65 and to calculate d we have:
$\begin{aligned} & d=-61-\left( -65 \right) \\\ & \Rightarrow d=-61+65 \\\ & \Rightarrow d=4 \\\ \end{aligned}$
So, now to get the $11^{th}$ term from the series of A.P, we can use the formula as:
${{a}_{11}}=a+\left( 11-1 \right)d$
Now, by substituting the value of a as -65 and value of d as 4 calculated above to get the value of ${{a}_{11}}$ as:
${{a}_{11}}=-65+\left( 11-1 \right)\times \left( 4 \right)$
So, now just by solving the above expression as:
$\begin{aligned} & {{a}_{11}}=-65+\left( 10 \right)\times \left( 4 \right) \\\ & \Rightarrow {{a}_{11}}=-65+\left( 40 \right) \\\ & \Rightarrow {{a}_{11}}=-25 \\\ \end{aligned}$
So, the $11^{th}$ term from the end of the A.P is $-25$.
Hence, the $11^{th}$ term from the end of the A.P is $-25$ as a final answer.

Note: In this type of questions, we must know the approach to solve as it is asked to find the $11^{th}$ term from the end and not from the beginning , that is why we reversed the entire A.P. If in a hurry we find the $11^{th}$ term from the beginning we get the wrong answer as:
$\begin{aligned} & {{a}_{11}}=27+\left( 11-1 \right)\times \left( -4 \right) \\\ & \Rightarrow {{a}_{11}}=27+\left( 10 \right)\times \left( -4 \right) \\\ & \Rightarrow {{a}_{11}}=27-40 \\\ & \Rightarrow {{a}_{11}}=-13 \\\ \end{aligned}$
Which is a wrong answer and we should take care while solving the same.