Math, asked by parthgarg2048, 4 months ago

Q. 7. Find value of p for which the consecutive terms 3k + 1, 2k + 3, 6k + 2 form an A.P.

Answers

Answered by ITZBFF
26

\mathsf\red{Given}

\mathsf{3k + 1, \: 2k + 3, \: 6k + 2 \: are \:  in \: A.P.}

\mathsf{}

\mathsf{To \: find \: the \: value \: of \: k : \:}

\mathsf\red{{2}^{nd} \: term \: - \: {1}^{st} \: term \: = \:{3}^{rd} \: term \: - \: {2}^{nd} \: term}

\mathsf{}

\mathsf{2k+3 \: - \: (3k+1) \: = \: 6k+2 \: - \: (2k+3)}

\mathsf{2k+3-3k-1 \: = \: 6k+2-2k-3}

\mathsf{-k+2 \: = \: 4k-1}

\mathsf{2+1 \: = \: 4k+k}

\mathsf{3 \: = \: 5k}

\mathsf{k \: = \: \frac{3}{5}} \\

\boxed{\mathsf\red{\therefore \: Value \: of \: k \: = \: \frac{3}{5}}}

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