Question

The sum of an A.P series is $k\left(\frac{a+c}{c-b}\right),$ whose last two terms are $b$ and $c$ respectively. If $a$ be the first term of the series, then the value of $k$ is

• A. $\frac{2c+b-a}{2}$
• B. $\frac{2c+b+a}{2}$
• C. $\frac{2c-b-a}{2}$
• D. $\frac{2c-b+a}{2}$

Answer 1

The correct option is C $\frac{2c-b-a}{2}$

$T_1=a,T_{n-1}=b,T_n=cd=T_n-T_{n-1}=c-bc=a+(n-1)d=a+(n-1)(c-b)\Rightarrow n=\frac{2c-b-a}{c-b}$

$S_n=k\left(\frac{a+c}{c-b}\right)\Rightarrow \frac{n}{2}(T_1+T_n)=k\left(\frac{a+c}{c-b}\right)\Rightarrow \left(\frac{2c-b-a}{2(c-b)}\right)\times (a+c)=k\left(\frac{a+c}{c-b}\right)\Rightarrow k=\frac{2c-b-a}{2}$