Find the sum of all the numbers between 100 and 200 which are divisible by 12.

Correct option is D

Given:
Range = 100 - 200
Formula Used:
S = $$\frac{n}2$$​×(a+l)
where,
S is the sum of the sequence,
n is the number of terms in the sequence,
a is the first term,
l is the last term.
Solution:
first and last numbers divisible by 12 between 100 and 200. The smallest multiple of 12 greater than 100 is 108.
The largest multiple of 12 less than 200 is 192.
finding the difference between the first and last terms, dividing by the common difference, and adding 1
Number of terms=192−$$\frac{108}{12}$$​ +1 = $$\frac{84}{12}$$​ +1 =7+1 =8
The formula for the sum S of an arithmetic sequence is
​$$S = \frac{n}2×(a+l) \\ \ \\S = \frac{8}2 \times (108 + 192)\\ \ \\S= 4 \times 300\\ \ \\S = 1200$$​​
The sum of all numbers between 100 and 200 that are divisible by 12 is ,
= 1200