Browse Questions

# The number of terms in the AP a, b,....c is

$(a)\;c\qquad(b)\;\frac{(c-a)}{(b-a)}\qquad(c)\;\frac{(b+c-2a)}{(b-a)}\qquad(d)\;\frac{(b+c+2a)}{(b-a)}$

Answer : (c) $\;\frac{(b+c-2a)}{(b-a)}$
Explanation : Let d be the common differance of the AP
Then d=b-a
The value of c is given by
$c=a+(n-1)d$
$c=a+(n-1)(b-a)$
Solving this , $c=2a-b\;+(b-a)n$
$n=\frac{(b+c-2a)}{(b-a)}.$