If the first term of an A.P. is a and nth term is b, then its common d

1.	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. \(\frac{b+a}{n-1}\)
Answer» Correct answer is B. \(\frac{(b-a)}{(n-1)}\) ‘a’ is the first term of the A.P, ‘b’ is the (n)^th term and ‘d’ is the common difference of the A.P. Then, we have b = a + (n–1) d = a + (n + 1)d Hence, ‘B’ = \(\frac{(b-a)}{(n-1)}\)

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. \(\frac{b+a}{n-1}\)

Answer»

Correct answer is B. \(\frac{(b-a)}{(n-1)}\)

‘a’ is the first term of the A.P,

‘b’ is the (n)^th term and ‘d’ is the common difference of the A.P.

Then, we have b = a + (n–1) d

= a + (n + 1)d

Hence, ‘B’ = \(\frac{(b-a)}{(n-1)}\)