Question

Find the arithmetic mean of the series given below:
$\text{1, 2, }{{\text{2}}^{\text{2}}},\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. , }{{\text{2}}^{n-1}}$

Answer 1

Hint: To find the arithmetic mean of any series, we first have to calculate the sum of the series. The sum of the series is given by adding all the terms of the series. Then, we find out the number of terms that are there in this series. The arithmetic mean is found by dividing the obtained sum by the number of terms in the series. The series that is given in the question is a geometric progression. Using this, we can solve this question.

Complete step-by-step answer:

Before proceeding with the question, we must know all the formulas that will be required to solve this question.

In sequences and series, if we are given a geometric progression having its first term equal to a, the common ratio equal to r and the number of terms equal to n, then the sum of this geometric progression is given by the formula,
$S=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ . . . . . . . . . . . . . . . (1)

The arithmetic mean of a series is given by dividing the sum of the series by the number of terms that are there in the series . . . . . . . . (2)

In the question, we are given a series $\text{1, 2, }{{\text{2}}^{\text{2}}},\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. , }{{\text{2}}^{n-1}}$ of which we are required to find the arithmetic mean.

It can be noticed that the number of terms in this series is n since the series is starting from ${{2}^{0}}$ and it is ending at ${{2}^{n-1}}$.

Also, this series is a geometric progression with first term as 1, common ratio as 2 and the number of terms equal to n. Using formula (1), the sum of this series is equal to,
$\begin{align}
  & S=\dfrac{1\left( {{2}^{n}}-1 \right)}{2-1} \\
 & \Rightarrow S={{2}^{n}}-1 \\
\end{align}$

Substituting the above two obtained numbers in formula (2), the arithmetic mean of this series is equal to, $\dfrac{{{2}^{n}}-1}{n}$


Note: There is a possibility that one may write the number of terms in the series as n-1 instead of n. But since the series is starting from ${{2}^{0}}$ and it is ending at ${{2}^{n-1}}$, the number of terms in the series is equal to n.