Question

Find the sum of the following series without actually adding it.
\[1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21\]

Answer 1

Hint: Here we will first observe that the given series is in Arithmetic progression and then we will use the formula for sum of series in AP.
The sum of series of AP with first term as a, last term as l, n as total number of terms as n is given by:-
\[sum = \dfrac{n}{2}\left[ {a + l} \right]\]

Complete step-by-step answer:
The given series is:-
\[1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21\]
The given series is in A.P.
Here, the first term \[a = 1\]
The last term is \[l = 21\]
Number of terms \[n = 11\]
Now we know that the sum of series of AP with first term as a, last term as l, n as total number of terms as n is given by:-
\[sum = \dfrac{n}{2}\left[ {a + l} \right]\]
Hence, putting in the values we get:-
\[sum = \dfrac{{11}}{2}\left[ {1 + 21} \right]\]
Solving it further we get:-
\[sum = \dfrac{{11}}{2}\left[ {22} \right]\]
\[
   \Rightarrow sum = 11 \times 11 \\
   \Rightarrow sum = 121 \\
 \]

Therefore, the sum of the given series is 121.

Note: The students can also use the following formula to find the sum of the given series.
\[sum = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
 Where, n is the total number of terms
a is the first term,
d is the common difference between two consecutive terms
Now in the given series, \[a = 1\], \[n = 11\], \[d = 2\]
Hence, putting in the respective values in the above formula we get:-
\[sum = \dfrac{{11}}{2}\left[ {2\left( 1 \right) + \left( {11 - 1} \right)2} \right]\]
Simplifying it further we get:-
\[sum = \dfrac{{11}}{2}\left[ {2 + \left( {10} \right)2} \right]\]
Solving it further we get:-
\[sum = \dfrac{{11}}{2}\left[ {22} \right]\]
\[
   \Rightarrow sum = 11 \times 11 \\
   \Rightarrow sum = 121 \\
 \]
Therefore, the sum of the given series is 121.
Students should also note that in an arithmetic progression the difference between two consecutive terms is always constant i.e. the common difference is always the same.