Question

Find the value of p for which the numbers 2p – 1, 3p + 1, 11 are in AP. Hence, find the numbers. ​

Answer 1

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Answer 2

$\sf \large \: Hint:-$

For solving this question we will directly use the basic concept of A.P

And the difference between the consecutive terms should be equal.

After that, we will see the condition for any three numbers a,b,c to be in A.P. and we will apply the result here to find the suitable value of p.

$\sf \large \: Solution \: \:: -$

The given three numbers (2p-1),(3p+1) and 11 are in AP.

Now,

$\sf2(3p + 1) = (2p - 1) + 11 \\ \sf \Rightarrow6p+2=2p-1+11 \\ \sf \Rightarrow6p+2=2p+10 \\ \sf \Rightarrow6p-2p=10-2 \\ \sf \: \Rightarrow4p=8 \\ \sf \: \Rightarrow p= \frac{8}{4} \\ \sf \Rightarrow p=2 \\ \bold { \red{ \underline{{the \: value \: of \: p \: is \: 2.}}}}$

Now,

the numbers will be

$\sf2 \times 2−1,3 \times 2+1,11 \\ \sf4−1,6+1,11 \\ \sf \red{\underline{{3,7,11}}}$