Question

Find the sums given below:

34 + 32 + 30 + … + 10

Answer 1

Given A.P: 34 + 32 + 30 + … + 10 

a = 34; d = a₂ – a₁ = 32 – 34 = -2 and the last term l = a_n = 10 

But, a_n = a + (n – 1) d 

∴ 10 = 34 + (n – 1) (-2) 

⇒ 10 – 34 = -2n + 2 

⇒ -2n = -24 – 2 

⇒ n = -26/-2 = 13

∴ n = 13 

Also, S_n = \(\frac{n}{2 }\) (a + l) 

where a = 34; l = 10 

S₁₃ = \(\frac{13}{2 }\)(34 + 10) 

= \(\frac{13}{2 }\) × 44 

= 13 × 22 

= 286