Question

If n arithmetic means are inserted between 2 and 38, then the sum of the resulting series obtained as 200. The value of n is
a. 6
b. 8
c. 9
d. 10

Answer 1

Hint: In order to solve this question, we should have some knowledge of arithmetic means, that is, if n arithmetic means are added between 2 numbers, then it will become an arithmetic progression of n + 2, where the first and the last terms are the given 2 numbers. Also, we need to know that for sum of n terms in AP, we can apply the formula, $S=\dfrac{n}{2}\left( a+l \right)$.

Complete step-by-step answer:

In this question, we have been given that n arithmetic means are inserted between 2 and 38 and the sum of the resulting series is 200. And we have been asked to find the value of n.
Let us consider the arithmetic mean inserted as ${{x}_{1}},{{x}_{2}},.......,{{x}_{n}}$. So, the series formed will be $2,{{x}_{1}},{{x}_{2}},{{x}_{3}},.......,{{x}_{n}},38$. And therefore, the number of terms of A in series will become n + 2. Now, we know that the arithmetic means of 2 terms form an AP with those numbers. So, to calculate the sum of (n + 2) terms, we can use the sum formula of AP, that is, $S=\dfrac{n}{2}\left( a+l \right)$. Where, n = n + 2, a = 2 and l = 38. So, we can write, $S=\dfrac{n+2}{2}\left( 2+38 \right)$.
Now, we have been given that the sum is 200. So, we get,
\[200=\dfrac{\left( n+2 \right)}{2}\left( 2+38 \right)\]
Now, we will simplify it, so we will get,
\[\begin{align}
  & 200=\dfrac{\left( n+2 \right)}{2}\left( 40 \right) \\
 & \Rightarrow 200=\left( n+2 \right)\left( 20 \right) \\
 & \Rightarrow \left( n+2 \right)=\dfrac{200}{20} \\
 & \Rightarrow \left( n+2 \right)=10 \\
 & \Rightarrow n=10-2 \\
 & \Rightarrow n=8 \\
\end{align}\]
Hence, we can say that 8 arithmetic means are inserted between 2 and 38. Therefore, option (b) is the correct answer.

Note: While solving this question, the possible mistakes we can make is in the sum formula, that is, we might think of using $S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$. But we do not have the value of d. So, we will get stuck somewhere. So, we will prefer to use $S=\dfrac{n}{2}\left( a+l \right)$, where l is the last term. Also, while calculating the sum, we might make a mistake by writing the number of terms as n instead of n + 2. So, we have to be careful while solving the question.