Question

!. Find the sum of 40 terms of an A.P whose third term is 1 and 6th term is -11.​

Answer 1

Given:-

a₃=1 and a₆=(-11)

To Find:-

S₄₀

Formula Applied:-

Sₙ=n/2[2a+(n-1)d]

Solution:-

a₃=a+2d

1=a+2d---------(1)

a₆=a+5d

(-11)=a+5d-----------(2)

Subtracting Equation(1) From Equation(2):-

1-(-11)=a-a+2d-5d

1+11=(-3)d

d=12/(-3)

d=(-4)

a+2d=1

a+2(-4)=1

a=1+8

a=9

S₄₀=40/2[2(9)+(20-1)(-4)]

S₄₀=20[18+(19)(-4)]

S₄₀=20(18-76)

S₄₀=20×(-58)

S₄₀=(-1160)

Answer 2

Answer: the correct answer is -2760

Step-by-step explanation: