Question

If the sum of the 12th and 22nd terms of an A.P. is 100, then the sum of the first 33 terms of the A.P. is

(a) 1650 

(b) 2340 

(c) 3300 

(d) 3400

Answer 1

Answer : (a) 1650

Let the first term of the A.P. be a and let the common difference be d.

Then, t₁₂ + t₂₂ = 100 

⇒ (a + 11d) + (a + 21d) = 100 

⇒ 2a + 32d = 100 ...(i) 

Now sum of first 33 terms of the A.P

= \(\frac{33}{2}\) (2a + 32d)  \(\big(\because S_n = \frac{n}{2}(2a+(n-1)d\big)\) 

= \(\frac{33}{2}\) × 100  (From (i)) 

= 1650.