Question

Determine k so that k^2+4k+8, 2k^2+3k+6, 3k^2+4k+4 are three consecutive terms of an ap.

Answer 1

Answer:

If a,b,c are three consecutive terms of an AP then 2b=a+c So here it is given that

${k}^{2} + 4k + 8$

,

$2 {k}^{2} + 3k + 6 \: and \: 3 {k}^{2} + 4k + 4$

are three consecutive terms of an AP.

$2(2 {k}^{2} + 3k + 6) = {k}^{2} + 4k + 8 +$

$3 {k}^{2} + 4k + 4 = >$

$4 {k}^{2} + 6k + 12 = 4 {k}^{2} + 8k + 12 = >$

$2k = 0 = > k = 0$

Answer 2

$\huge\fcolorbox{blue}{red}{Answer = 0}$

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$\bf\huge\underline{Explaination:-}$⠀

Since, k² + 4k + 8, 2k² + 3k + 6, and⠀⠀ 3k² + 4k + 4 are consecutive terms of an AP.

Therefore, 2k² + 3k + 6 - (k² + 4k + 8)

= 3k² + 4k + 4 - (2k² + 3k + 6)

= Common Difference

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=> 2k² + 3k + 6 - k² - 4k - 8

=> 3k² + 4k + 4 - 2k² - 3k - 6

=> k² - k - 2 = k² + k - 2

=> -k² = k

=> 2k = 0

=> k = 0

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$<marquee>Thank You$⠀