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Determine k, so that k^(2)+4k+8, 2k^(2)+...

Determine `k`, so that `k^(2)+4k+8, 2k^(2)+3k+6` and `3k^(2)+4k+4` are three consecutive terms of an AP.

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To determine the value of \( k \) such that the expressions \( k^2 + 4k + 8 \), \( 2k^2 + 3k + 6 \), and \( 3k^2 + 4k + 4 \) are three consecutive terms of an Arithmetic Progression (AP), we can follow these steps: ### Step 1: Define the terms Let: - \( A_1 = k^2 + 4k + 8 \) - \( A_2 = 2k^2 + 3k + 6 \) - \( A_3 = 3k^2 + 4k + 4 \) ...
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Knowledge Check

  • Assertion: If the terms k^(2) + 4k+8, 2K^(2)+ 3k +6 and 3K^(2) + 4k + 4 are in A.P., then the value of k is 0. Reason: If a, b, care in A.P., then a +b=b+c.

    A
    Both assertion and reason are correct and reason is the correct explanation of assertion.
    B
    Both assertion and reason are correct but reason is not the correct explanation of assertion.
    C
    Assertion is correct but reason is incorrect.
    D
    Assertion is incorrect but reason is correct.

  • If x^(2)+y^(2)=k^(2) , and xy=8-4k , what is (x+y)^(2) in terms of k ?

    A
    `k-4`
    B
    `(k-4)^(2)`
    C
    `k^(2)-4k+8`
    D
    `(k-2)^(2)+4`

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