Question

Find the sum of 60 terms of A.P 3,5,7,…

Answer 1

Hint: To find sum of n terms in an AP, the following formula is used:
     \[{{S}_{n}}=\dfrac{n}{2}\left( \text{first term}+\text{last term} \right)\]
To find the nth term, the following formula is used:
     \[{{T}_{n}}=a+\left( n-1 \right)d\]
Where, a = first term of the series.
     n = number of terms in the series.
     d = common difference of the series.

Complete step-by-step answer:
Since, the nth term is not given in the question, nth term is calculated. nth term is given by the formula:
     \[{{T}_{n}}=a+\left( n-1 \right)d\]
Here,
a = first term of the series = 3
     n = number of terms in the series = 60
d = common difference of the series = \[5-3=2\]
Substituting these values in the formula, we get,
     \[\begin{align}
  & {{T}_{n}}=a+\left( n-1 \right)d \\
 & {{T}_{n}}=3+\left( 60-1 \right)2 \\
 & {{T}_{n}}=3+\left( 59 \right)2 \\
 & {{T}_{n}}=3+118 \\
 & {{T}_{n}}=121 \\
\end{align}\]
Therefore, last term = 121
Now, we can substitute the value of last term in the formula to find sum of n terms in an AP as follows:
      \[\begin{align}
  & {{S}_{n}}=\dfrac{n}{2}\left( \text{first term}+\text{last term} \right) \\
 & {{S}_{n}}=\dfrac{60}{2}\left( 3+121 \right) \\
 & {{S}_{n}}=\dfrac{60}{2}\left( 124 \right) \\
 & {{S}_{n}}=30\left( 124 \right) \\
 & {{S}_{n}}=3720 \\
\end{align}\]
Therefore, the sum of 60 terms of the given AP is 3720.

Note: The formula for the nth term of an AP and sum of n terms of an AP should be remembered. The common difference between any two terms of an AP is always same. Hence, common difference can be calculated by subtracting any two consecutive numbers, keeping in mind that any number in the series is subtracted from the number succeeding it. Since, this is a series of odd numbers, if the series started from 1, sum of the series could have been simply calculated as \[{{n}^{2}}\]. Another formula for directly finding sum of the given series is:
          \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]