Question

Find the \[{{31}^{st}}\]term of an A.P series.

Answer 1

Hint: : Find the A.P of the numbers. To find the A.P of the number we will use the conventional formula and calculate the \[{{31}^{st}}\] term of an A.P series.
The first term and last term are 7 and 125 respectively and the total number of terms in a series is 60.

Complete step-by-step answer:
The formula for A.P.
\[\begin{align}
  & \text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }{{\text{a}}_{n}}={{a}_{1}}+\left( n-1 \right)d \\
 & {{\text{a}}_{n}}=\text{ the term }{{\text{n}}^{th}}\text{in the sequence} \\
 & {{a}_{1}}=\text{ the first term in the sequence} \\
 & d=\text{ the common difference between terms} \\
\end{align}\]
First step will be putting the given first and last terms.
We will find the common difference between the two consecutive terms.
We will use the common difference and first calculate the \[{{31}^{st}}\]term of an A.P series.
\[\begin{align}
  & {{\text{a}}_{n}}={{a}_{1}}+\left( n-1 \right)d \\
 & {{\text{a}}_{60}}={{a}_{1}}+\left( n-1 \right)d \\
 & 125=7+\left( 60-1 \right)d \\
 & d\cdot 59=125-7 \\
 & d=\dfrac{118}{59} \\
 & d=2 \\
 & \therefore \text{ }{{\text{a}}_{n}}={{a}_{1}}+\left( n-1 \right)d \\
 & {{\text{a}}_{31}}={{a}_{1}}+\left( n-1 \right)d \\
 & {{\text{a}}_{31}}=7+\left( 31-1 \right)2 \\
 & {{\text{a}}_{31}}=7+30\cdot 2 \\
 & {{\text{a}}_{31}}=67 \\
\end{align}\]
Thus, the \[{{31}^{st}}\]term of an A.P series is 67.

Note: A mathematical sequence in which the difference between two consecutive terms is always constant and it is called Arithmetic Progression.