Question

What is the sum of the first 12 odd numbers $1,3,5,7,...$?
(a) 12
(b) 144
(c) 141
(d) 124

Answer 1

Hint:Observe that the given sequence is an AP with 1 being the first term. Calculate the common difference of this AP by subtracting any two consecutive terms. To calculate the sum of ‘n’ terms of AP, use the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where ‘a’ is the first term of the AP and ‘d’ is the common difference.

Complete step-by-step answer:
We have to calculate the sum of the first 12 odd numbers $1,3,5,7,...$.
We observe that the sequence of odd numbers form an AP, with 1 being the first term.
We will now calculate the common difference of this AP. To do so, we will subtract any two consecutive terms. Thus, the common difference is $=3-1=2$.
We will now calculate the sum of the first 12 odd numbers. We know that formula for calculating the sum of ‘n’ terms of AP ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where ‘a’ is the first term of the AP and ‘d’ is the common difference.
Substituting $a=1,d=2,n=12$ in the above formula, we have ${{S}_{12}}=\dfrac{12}{2}\left[ 2\left( 1 \right)+\left( 12-1 \right)2 \right]$.
Simplifying the above expression, we have ${{S}_{12}}=\dfrac{12}{2}\left[ 2\left( 1 \right)+\left( 12-1 \right)2 \right]=6\left( 2+11\left( 2 \right) \right)=6\left( 2+22 \right)=6\times 24=144$.
Hence, the sum of the first 12 odd numbers is 144, which is option (b).

Note: We can also solve this question by calculating the ${{12}^{th}}$ term of the AP using the formula ${{a}_{n}}$=$a+\left( n-1 \right)d$, where ${{a}_{n}}$ represents the ${{n}^{th}}$ term of an A.P and then use the formula ${{S}_{n}}=\dfrac{n}{2}\left[ a+{{a}_{n}} \right]$ to calculate the sum of ‘n’ terms. We can also write the first 12 odd numbers and add them up to calculate the sum. However, it will be very time-consuming.