Question

for an ap a is equal to 11 and d equal to 7 find the sum of the first 40th
terms of AP​

Answer 1

Answer:

S₄₀ = 5900

Step-by-step explanation:

given an AP with a=11;d=7

∴ S₄₀ = $\frac{40}{2}$ [2(11) + (40-1)7 ]           [ ∵ Sₙ = $\frac{n}{2}$(2a+(n-1)d) ]

         = 20 [ 22+39(7) ]

         = 20 (22+273)

         = 20(295)

         = 5900