Question 11
Determine k, so that $k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6 a n d 3 k^{2} + 4 k + 4$ are three consecutive terms of an AP.

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Solution

Given, $k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6 a n d 3 k^{2} + 4 k + 4$ are consecutive terms of an AP.
$∴ 2 k^{2} + 3 k + 6 - (k^{2} + 4 k + 8) = 3 k^{2} + 4 k + 4 - (2 k^{2} + 3 k + 6)$ = common difference
$\Rightarrow 2 k^{2} + 3 k + 6 - k^{2} - 4 k - 8 = 3 k^{2} + 4 k + 4 - 2 k^{2} - 3 k - 6$
$\Rightarrow k^{2} - k - 2 = k^{2} + k - 2$
$\Rightarrow - k = k \Rightarrow 2 k = 0 \Rightarrow k = 0$

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K

K^{2} + 4 K + 8, 2 K^{2} + 3 K + 6

3 K^{2} + 4 K + 4

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k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6, 3 k^{2} + 4 k + 4

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k^{2} + 4 k + 8

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k^{2} + 4 k + 8 = 0, 2 k^{2} + 3 k + 6 = 0, 3 k^{2} + 4 k + 4 = 0

Determine the value of k for which $k^{2} + 4 k + 8, 2 k^{2} + 3 k + 6$ and $3 K^{2} + 4 k + 4$ are in A.P.

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