Question

For what value of k, are the numbers p, 2p+k and 3p+6 three consecutive terms of AP?​

Answer 1

$\textbf{Given:}$

$\textsf{p, 2p+k, 3p+6 are in A.P}$

$\textbf{To find:}$

$\textsf{The value of k}$

$\textbf{Solution:}$

$\textbf{Concept used:}$

$\boxed{\begin{minipage}{5cm} \\\mathsf{\;\;For\;any\;A.P,\;\;t_2-t_1=t_3-t_2}\\ \end{minipage}}$

$\textsf{Since p, 2p+k, 3p+6 are in A.P,}$

$\mathsf{t_2-t_1=t_3-t_2}$

$\implies\mathsf{2p+k-p=(3p+6)-(2p+k)}$

$\implies\mathsf{p+k=3p+6-2p-k}$

$\implies\mathsf{p+k=p+6-k}$

$\implies\mathsf{k+k=6}$

$\implies\mathsf{2k=6}$

$\implies\boxed{\mathsf{k=3}}$

$\textbf{Find more:}$

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Answer 2

SOLUTION

TO DETERMINE

The value of k for which the numbers p, 2p+k and 3p+6 three consecutive terms of AP

FORMULA TO BE IMPLEMENTED

If three numbers a, b, c are in AP ( Arithmetic progression) then

$\sf{2b = a + c}$

EVALUATION

Here it is given that the numbers p, 2p+k and 3p+6 three consecutive terms of AP

$\therefore \sf{ \: \: 2(2p + k) =p + (3p + 6) }$

$\sf{ \implies\: 4p + 2k=4p + 6 }$

$\sf{ \implies\: 2k=6 }$

$\sf{ \implies\: k=3}$

FINAL ANSWER

The required value of k is 3

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