Question

Find the sum of first 40 terms of an AP whose 4th term is 8 and 6th term is 14

Answer 1

a4=8

an=a+(n-1)×d

8=a+(4-1)×d

8=a+3d

a6=14

an=a+(n-1)×d

14=a+(6-1)×d

14=a+5d

a+3d=8.........1eq

a+5d=14..........2eq

- - -

__________

-2d=-6

d=6÷2=3

put the value of d in eq1

a+3d=8

a+3×3=8

a+9=8

a=9-8

a=1

Sn=n÷2(2a+(n-1)×d

S40=40÷2(2×1+(40-1)×3

S40=20(2+39×3)

S40=20(2+117)

S40=20(119)

S40=2380