Question

What is the sum of first 14 terms of the A.P. $12,15,18,\mathrm{21....}?$

• A. $144$
• B. $114$
• C. $441$
• D. $414$

Answer 1

The correct option is C $441$

First term of the given arithmetic series = 12
Second term of the given arithmetic series = 15
Third term of the given arithmetic series = 18
Fourth term of the given arithmetic series = 21
Now, Second term - First term = 15 - 12 = 3
Third term - Second term = 18 - 15 = 3
Therefore, common difference of the given arithmetic series is 3.
The number of terms of the given A. P. series (n) = 14
We know that the sum of first n terms of the Arithmetic Progress, whose first term = a and common difference = d is given by:
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$S_{14}=\frac{14}{2}[2\times 12+(14-1)\times 3]$
$S_{14}=7[24+39]$
$S_{14}=441$