6 + 12 + 18 + 24 + ....... +612 = ?

Correct option is A

Given:
Series: 6 + 12 + 18 + 24 + ....... + 612
Formula Used:
Sum of an Arithmetic Progression (A.P.):
​$$S_n = \frac{n}{2}(a + l)$$​​
Number of terms n = $$\frac{l - a}{d} + 1$$​​
Solution:
The given series is an A.P. with:
First term a = 6
Common difference d = 6
Last term l = 612
Calculate the total number of terms n:
n = $$\frac{612 - 6}{6} + 1$$​​
n = $$\frac{606}{6} + 1$$​​
n = 101 + 1
n = 102
Now, calculate the sum of the series:
​$$S_{102} = \frac{102}{2} × (6 + 612)$$​​
 = 51 × 618
 = 31518
Final Answer
So the correct answer is (a)