Question

What is the ${30^{th}}$ term of arithmetic series $11 + 16 + 21 + 26 + ....$?

Answer 1

Hint: Here we are finding the ${30^{th}}$ term of the given arithmetic series by using the common difference between the terms and also using the formula for finding ${n^{th}}$ term of the arithmetic series.
Formula used:
Finding ${n^{th}}$ term of the arithmetic series using ${T_n} = {a_1} + \left( {n - 1} \right)d$
Where ${a_1}$ is the first term of the series and $d$ is the common difference.
To calculate the common difference of a arithmetic series , difference between the second term of the series and the first term or simply find the difference of any two consecutive terms by taking the previous term , that is $d = {a_2} - {a_1}$ where ${a_2}$ is the second term of the series.

Complete step-by-step solution:
Given arithmetic series $11 + 16 + 21 + 26 + ....$
First term of the series is ${a_1} = 11$ and the common difference of the series is $d = {a_2} - {a_1} = 16 - 11 = 5$.
Now finding${n^{th}}$ term of the arithmetic series using ${T_n} = {a_1} + \left( {n - 1} \right)d$
Substitute in the values of ${a_1} = 11$ and $d = 5$ we get,
${T_n} = 11 + \left( {n - 1} \right)5$
Now we are going to find the ${30^{^{th}}}$ term of the series ,
Substitute in the value of $n$ to find the$n$ the term, that is $n = 30$we get,
${T_{30}} = 11 + \left( {30 - 1} \right)5$
Now simplify the bracket term first we get,
${T_{30}} = 11 + \left( {29} \right)5$
Multiply the term and we get,
${T_{30}} = 11 + 145$
And add the term we get,
${T_{30}} = 156$

Note: An arithmetic series is the sum of the terms of an arithmetic sequence. A geometric series is the sum of the terms of a geometric sequence. We can only take partial sum of an arithmetic sequence. The partial sum is the sum of a limited number of terms, like the first ten terms, or the fifth through the hundredth terms.