Question

Find k, such that ${k^2} + 4k + 8,2{k^2} + 3k + 6,3{k^2} + 4k + 4$ are 3 consecutive terms of an A.P.

Answer 1

Hint: A.P. is the short form of Arithmetic progression. An arithmetic progression is a sequence of numbers such that the difference between any two consecutive numbers is a constant and this constant is termed a common difference. If 3 consecutive terms of an A.P. are given, then twice of the middle term is equal to the sum of the first and third terms. Use this to find the value of k.

Complete step-by-step answer:
We are given that ${k^2} + 4k + 8,2{k^2} + 3k + 6,3{k^2} + 4k + 4$ are 3 consecutive terms of an A.P.
We have to find the value of k.
Here ${k^2} + 4k + 8$ is the first term, $2{k^2} + 3k + 6$ is the second term and $3{k^2} + 4k + 4$ is the third term
Let the first term be A, second term be B and third term be C and A, B and C are consecutive terms in an Arithmetic Progression.
We know that twice of the middle term is equal to the sum of the first and third terms, $2B = A + C$
Substitute the values of A, B and C.
$
2 \times \left( {2{k^2} + 3k + 6} \right) = \left( {{k^2} + 4k + 8} \right) + \left( {3{k^2} + 4k + 4} \right) \\
\to 4{k^2} + 6k + 12 = {k^2} + 4k + 8 + 3{k^2} + 4k + 4 \\
 $
Putting all the similar terms together
$
 4{k^2} + 6k + 12 = {k^2} + 3{k^2} + 4k + 4k + 8 + 4 \\
 \to 4{k^2} + 6k + 12 = 4{k^2} + 8k + 12 \\
 $
On cancelling the similar terms on both sides, we get
$6k = 8k$
The above equation satisfies only when the value of k is zero.
Therefore, the value of k is 0.
The terms are $8,6,4$

Note: Arithmetic sequence is a sequence of numbers such that the difference of any two successive members is a constant which is called common difference whereas in a geometric progression, every successive term is obtained by multiplying the previous term with a fixed number known as common ratio. The difference we can see is addition of common difference and multiplication of common ratio in these two progressions. So be careful while finding the unknown terms of a specific progression.