Question

For what value of k: 2k, \[k+10\] and \[3k+2\] are in A.P?

Answer 1

Hint: In this case, we will use the conditions for three consecutive terms to be, which is \[2b=a+c\] where b is the middle term, a is the first term and c is the third term. We will receive one variable equation here, which is k.

Complete step-by-step answer:
Given, 2k, \[k+10\] and \[3k+2\] are three consecutive terms of an A.P
As we know that for the three consecutive terms a, b and c to be in A.P, twice of the second term should be equal to sum of first and last that is \[2b=a+c\]
For the given terms to be in an A.P, twice of the middle term which is\[k+10\]should be equal to sum of the first term which is 2k and second term which is \[3k+2\]
\[2\left( k+10 \right)=\left( 2k \right)+\left( 3k+2 \right)\]
By simplifying this we get:
\[2k+20=5k+2\]
By rearranging the term, we get:
\[18=3k\]
Therefore, we get the value of k from above equation:
\[k=6\]
Hence, the required value of k is 6.
So, the correct answer is “\[k=6\]”.

Note: Arithmetic progression (AP) is a set of terms in which the common difference between them remains the same. The difference between two words should be equal to the difference between the last two terms for any three consecutive terms a, b, and c to be in AP that is \[b-a=c-b\] from this we get the expression of Arithmetic progression is given as \[2b=a+c\].