Question

Prove that no matter what the real numbers $a$ and $b$ are, the sequence with $n$th term $a+nb$ is always an A.P. What is the common difference?

Answer 1

Given the sequence which is defined by.

$a_n=a+nb.....eq\left(1\right)$

and we know that nth term of an A.p is given by

$a_n=a_1+(n-1)d$

which can also be written as

$a_n=a_1-d+nd.....eq\left(2\right)$

comparing eq(1) and eq(2) we get

common difference is

$d=b\left(bycomparingcoefficientsofn\right)$

and

$a_1-d=a$

by putting value of d=b in above equation we get

$a_1-b=a$

$⟹a_1=a+b$ (first term of A.P)

hence if a and b are real numbers this sequence form an A.P with first term $a_1=a+b$

and have a common difference $d=b$