Question

if the sum of first n terms of an ap is 4n-n^2 then first fourth term is​

Answer 1

Answer:

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

S₉=4(9)-(9)²= 36–81 = -45

S₁₀=4(10)-(10)²= 40–100 = -60

Alternatively, using eqn (1), with a = 3 and d= -2

S₉= (9/2)[2(3)+(9-1)(-2)]= -45

S₁₀= (10/2)[2(3)+(10–1)(-2)]= -60

Now, 10 th term, T₁₀=S₁₀-S₉ = -60–(-45)=-15

nth term, Tₙ = a+(n-1) d= 3+(n-1)(-2)= 5–2n

Ans: 3rd term =-1; 10th term = -15 ; nth term = (5–2n)

Answer 2

answer is zero

we should multiipy 4 in n value place so we zero as final ans