Question:
If the first term of an A.P. is a and nth term is b, then its common difference is
(a) $\frac{b-a}{n+1}$
(b) $\frac{b-a}{n-1}$
(C) $\frac{b-a}{n}$
(d) $\frac{b+a}{n-1}$
Solution:
Here, we are given the first term of the A.P. as a and the nth term (an) as b. So, let us take the common difference of the A.P. as d.
Now, as we know,
$a_{n}=a+(n-1) d$
On substituting the values given in the question, we get.
$b=a+(n-1) d$
$(n-1) d=b-a$
$d=\frac{b-a}{n-1}$
Therefore, $d=\frac{b-a}{n-1}$
Hence the correct option is (b).