Question

solve this please

If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.​

Answer 1

Step-by-step explanation:

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)