Question

Find the sum of the following AP.
$\dfrac{1}{15},\dfrac{1}{12},\dfrac{1}{10},....$ to 11 terms.

Answer 1

Hint: The sum of an arithmetic progression is given by the formula,
${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ , where ${{S}_{n}}$ is the sum, n is the number of terms in the progression, a is the first term of the progression and d is the common difference between any two consecutive terms in the progression. So, we substitute the values from the given progression and find the sum up to 11 terms.

Complete step by step solution:
The given question is based on sequences and series, particularly arithmetic progression. The sequence can be called as an arithmetic progression if and only if the common difference between every two consecutive terms is equal. In the question, we are given the series,
$\dfrac{1}{15},\dfrac{1}{12},\dfrac{1}{10},....$to 11 terms.
Now, let us check the common difference d between the first 2 consecutive terms.
$\begin{align}
 & d=\dfrac{1}{12}-\dfrac{1}{15} \\
 & \Rightarrow d=\dfrac{15-12}{\left( 15\times 12 \right)} \\
 & \Rightarrow d=\dfrac{3}{180} \\
 & \Rightarrow d=\dfrac{1}{60} \\
\end{align}$
For the second and third term, we get
$\begin{align}
 & d=\dfrac{1}{10}-\dfrac{1}{12} \\
 & \Rightarrow d=\dfrac{12-10}{\left( 10\times 12 \right)} \\
 & \Rightarrow d=\dfrac{2}{120} \\
 & \Rightarrow d=\dfrac{1}{60} \\
\end{align}$
Hence, the common difference is constant and now we can move on to finding the sum of the progression. By inspection, we can clearly see that the first term $a=\dfrac{1}{15}$ . The number of terms in the progression is given as 11. Therefore, we can now find the sum of the progression as,
$\begin{align}
 & {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \left( 2\times \dfrac{1}{15} \right)+\left( 11-1 \right)\dfrac{1}{60} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{2}{15}+\dfrac{10}{60} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{2}{15}+\dfrac{1}{6} \right) \\
\end{align}$
Now, taking the LCM and simplifying further, we get
$\begin{align}
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{\left( 2\times 6 \right)+\left( 1\times 15 \right)}{15\times 6} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{12+15}{90} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{27}{90} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{3}{10} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{33}{20} \\
\end{align}$
Hence, the sum of the given arithmetic progression is ${{S}_{n}}=\dfrac{33}{20}$.

Note: We can also find the sum of the progression by finding the last term of the progression. We know that at every term, the common difference gets added so we can use the last term to find the sum by using the below formula,
${{S}_{n}}=\dfrac{n}{2}\left( a+{{t}_{n}} \right)$ , where ${{S}_{n}}$ is the sum, n is the number of terms in the progression, a is the first term of the progression and ${{t}_{n}}$ is the last term. We can find the last term as follows,
$\begin{align}
 & {{t}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow {{t}_{n}}=\dfrac{1}{15}+\left( 11-1 \right)\dfrac{1}{60} \\
 & \Rightarrow {{t}_{n}}=\dfrac{1}{15}+\dfrac{10}{60} \\
 & \Rightarrow {{t}_{n}}=\dfrac{1}{15}+\dfrac{1}{6} \\
 & \Rightarrow {{t}_{n}}=\dfrac{6+15}{90} \\
 & \Rightarrow {{t}_{n}}=\dfrac{21}{90} \\
 & \Rightarrow {{t}_{n}}=\dfrac{7}{30} \\
\end{align}$
Using this, we can find sum as,
$\begin{align}
 & {{S}_{n}}=\dfrac{n}{2}\left( a+{{t}_{n}} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{1}{15}+\dfrac{7}{30} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{2+7}{30} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{1}{15}+\dfrac{7}{30} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{9}{30} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{11}{2}\left( \dfrac{3}{10} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{33}{20} \\
\end{align}$
Hence, the sum of the progression is ${{S}_{n}}=\dfrac{33}{20}$.