Question

3
Find the sums given below:
(1) 34 + 32 + 30 + ... + 10​

Answer 1

34 + 32 + 30 + ... + 10

the given series is in A.P with common difference of -2.

• so here a = 34 ; d = -2 ; aₙ = 10

         aₙ = a + (n-1)d

      ⇒ 10 = 34 + (n-1)-2

       ⇒ 2(n-1) = 24

       ⇒  n = 11 - 1 = 10

• so sum of n terms of an AP is given by :

Sₙ = n/2 [2a + (n – 1)d]

        ⇒ Sₙ = 10/2 [ 2 × 34 + (10-1)×-2]

                  Sₙ = 250

Hope this helps you ✌️✌️☘️☘️.