Question

Find the value of k, if k2 + 4k + 8, 2k2 + 3k + 4 and 3k2 + 4k + 4 are 3 consecutive terms of an AP.

Answer 1

Answer:-

$\sf{The \ value \ of \ k \ is \ -2.}$

Given:

• $\sf{Three \ consecutive \ terms \ of \ AP \ are}$
• $\sf{k^{2}+4k+8, \ 2k^{2}+3k+4 \ and }$
• $\sf{3k{2}+4k+4.}$

To find:

• The value of k.

Solution:

$\sf{Let \ t_{1}=k^{2}+4k+8,}$

$\sf{t_{2}=2k^{2}+3k+4}$

$\sf{t_{3}=3k^{2}+4k+4}$

$\boxed{\sf{2\times \ t_{2}=t_{1}+t_{3}}}$

$\sf{\therefore{2(2k^{2}+3k+4)=(k^{2}+4k+8)+(3k^{2}+4k+4)}}$

$\sf{\therefore{4k^{2}+6k+8=4k^{2}+8k+12}}$

$\sf{\therefore{6k-8k=12-8}}$

$\sf{\therefore{-2k=4}}$

$\sf{\therefore{k=-\frac{4}{2}}}$

$\boxed{\sf{\therefore{k=-2}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ k \ is \ -2.}}}$

Answer 2

Given that ,

The three consecutive terms of AP is k² + 4k + 8 , 2k² + 3k + 4 , and 3k² + 4k + 4

We know that , the difference of two consecutive terms of an AP is constant or same

Thus ,

$\rm \hookrightarrow 2 {k}^{2} + 3k + 4 - ( {k}^{2} + 4k + 8) = 3 {k}^{2} + 4k + 4 - (2 {k}^{2} + 3k + 4)</p><p> \\ \\ \rm \hookrightarrow</p><p>2 {k}^{2} + 3k + 4 - {k}^{2} - 4k - 8 = 3 {k}^{2} + 4k + 4 - 2 {k}^{2} - 3k - 4 \\ \\ \rm \hookrightarrow {k}^{2} - k - 4 = {k}^{2} + k \\ \\ \rm \hookrightarrow - 2k = 4 \\ \\ \rm \hookrightarrow k = - 2$

$\therefore \rm \underline{The \: value \: of \: k \: is -2}$