Arithmetic progression

Question

The 16th term of an arithmetic series is seven times the second term.

Show that the 13th term is four times the third term.

Answer 1

a_n  = nth term = a₁ + (n-1)d

 

Since a₁₆ = 7a₂, a₁ + 15d = 7(a₁ + d)

 

                             a₁ + 15d = 7a₁ + 7d

 

                                      6a₁ = 8d     So, a₁ = (4/3)d

 

a₁₃ = a₁ + 12d = (4/3)d + 12d = (40/3)d

 

a₃ = a₁ + 2d = (4/3)d + 2d = (10/3)d

 

(40/3)d = 4(10/3)d

 

∴ a₁₃ = 4a₃