Question

If $k+2,4k-6,3k-2$ are the three consecutive terms of an A.P., then the value of $k$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (B)

$3$

It is given that $k+2,4k-6,3k-2$ are the consecutive terms of A.P, therefore, by arithmetic mean property we have, first term+third term is equal to twice of second term that is:

$(k+2)+(3k-2)=2(4k-6)\Rightarrow 4k=8k-12\Rightarrow 4k-8k=-12\Rightarrow -4k=-12\Rightarrow 4k=12\Rightarrow k=\frac{12}{4}=3$

Hence $k=3$.