Question

Find the A.P, if the ${10}^{th}$ term of an A.P. is 52 and the ${17}^{th}$ term of an A.P. is 20 more than the ${13}^{th}$ term.

Answer 1

Hint: We will use the ${n}^{th}$ term formula \[{{T}_{n}}=a+(n-1)d\] of arithmetic progression a few times to solve this question. As the ${10}^{th}$ term is given we will apply the above formula and then we will get a relationship between a and d. Also we will substitute the ${13}^{th}$ term in ${17}^{th}$ term formula to get the common difference.

Complete step-by-step answer:
Before proceeding with the question we should understand the concept of arithmetic progression.
Arithmetic Progression (AP) is a sequence of numbers in order that the difference of any two successive numbers is a constant value. For example, the series of natural numbers: 1,2,3,4,5, 6, … is an AP, which has a common difference between two successive terms (say 1 and 2) equal to 1 (2 -1). Even in the case of odd numbers and even numbers we can see the common difference between two successive terms will be equal to 2.
We know that the ${n}^{th}$ term formula of an AP is \[{{T}_{n}}=a+(n-1)d......(1)\] where a is the first term and d is the common difference.
It is mentioned in the question that the ${10}^{th}$ term is 52. So substituting n equal to 10 in equation (1) we get,
\[\begin{align}
  & {{T}_{10}}=a+(10-1)d \\
 & 52=a+9d...........(2) \\
\end{align}\]
Also it is mentioned that the ${17}^{th}$ term is 20 more than the ${13}^{th}$ term. So using this information in equation (1) we get,
\[\begin{align}
  & {{T}_{17}}=a+(17-1)d \\
 & {{T}_{13}}+20=a+16d...........(3) \\
\end{align}\]
Now finding the ${13}^{th}$ term using equation (1) we get,
\[\begin{align}
  & {{T}_{13}}=a+(13-1)d \\
 & {{T}_{13}}=a+12d...........(4) \\
\end{align}\]
Now substituting the value of ${13}^{th}$ term from equation (4) in equation (3) we get,
\[a+12d+20=a+16d...........(5)\]
Now cancelling the similar terms in equation (5), we get,
\[\begin{align}
  & 16d-12d=20 \\
 & 4d=20 \\
 & d=\dfrac{20}{4}=5......(6) \\
\end{align}\]
Now substituting the value of d from equation (6) in equation (2) we get,
\[\begin{align}
  & 52=a+9\times 5 \\
 & a=52-45=7 \\
\end{align}\]
So we have got the first term as 7 and the common difference as 5. Now according to the definition the A.P is 7, 12, 17, 22, 27, 32,……….

Note: Remembering the formula of the ${n}^{th}$ term of an arithmetic progression is the key here. We have to be careful while solving equation (3) and equation (4) because the ${17}^{th}$ term is 20 more than the ${13}^{th}$ and hence understanding this statement is important.