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.