Question

please answer fast ................​

Answer 1

For any AP, sum upto first n terms,

Sn= (n/2)[2a+(n-1)d] ——(1)

For given AP, sum of the first n terms, Sn= 4n-n²

First term, a = T₁ =S₁= 4(1)-(1)² = 4–1=3

Sum of first 2 terms, S₂ = 4(2)-(2)² = 8–4=4

Second term, T₂= S₂-S₁ = 4–3 =1

Sum of first 3 terms, S₃ = 4(3)-(3)² =12–9=3

3rd term, T₃=S₃-S₂ = 3–4=-1

Therefore, the series is: 3, 1, -1

Common difference, d =T₃-T₂= T₂-T₁= -2

hope it helps