Q. If the sum of 12th and 22nd terms of an AP is 100, then the sum of the first 33 terms of the AP is
A) $1700$
B) 1650
C) 3300
D) 3400
Explanation
(b)
Here, $T_{12} = a + 11d$
and $T_{22} = a + 21d$
Since, $100 = T_{12} + T_{22}$
$\therefore$ $100 = a + 11d + a + 21d$
$\Rightarrow \ \ a + 16d = 50$ …(i)
Now, $S_{33} = \frac{33}{2}\lbrack 2a + (33 - 1)d\rbrack$
$= 33(a + 16d)$
$= 33 \times 50 = 1650$ [from Eq. (i)]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.