Question

How do you give a recursive formula for the arithmetic sequence where the 4th term is 3; 20th term is 35?

Answer 1

given ,a4=3, a20=35

a+3d=3 (i)

a+19d=35 (ii)

equating both the equation,

a is cancelled out and d=(32/16)

d=2

now put the value of d in equation(i)

a+3d=3

a+3×2=3

a=3-6

a=(-3),d=2

so that recursive formula for AP. is

a1=(-3)

a2=a+d=(-3+2)=-1

a3=a+2d=(-3+2×2)=1

so the AP is (-3,-1,1)