Question

show that the sequence defined by an=4n+4 is an A.P. also, find its common difference.

Answer 1

given

an = 4n + 4

a1 = 4*1+4 = 8

a2 = 4*2+4 = 12

A3 = 4*3+4 = 16

a3 - a2 = 16 - 12 = 4

a2 - a1 = 12 - 8 = 4

now

ar = 4r + 4

a(r+1) = 4(r+1) + 4

4r + 8

a(r+1) - ar = 4r + 8 - 4r - 4

= 4

thus the sequence is AP with first term as 8 and common difference as 4.

Answer 2

Answer:

$a_{n}$ = 4n + 4

$a_{n-1}$ = 4(n-1) + 4

       = 4n -4 +4

       = 4n

Subtracting $a_{n-1}$ from $a_{n}$

   4n + 4 - (4n)

    4n +4 -4n

         4

Therefore, The sequence is a part of an AP

∵ $a_{n} - a_{n-1}$ is free from n

Common difference = $a_{n} - a_{n-1}$ = 4