Question

if (p-1) (p+3) (3p-1) are in AP than find the value of P

Answer 1

If ( p - 1 ) ( p+3) (3p-1) are consecutive pair of An AP then ,

$\frac{(p - 1) + (3p - 1)}{2} = (p + 3) \\ \\ \frac{4p - 2}{2} = p + 3 \\ \\ \frac{2(2p - 1)}{2} = p + 3$


$2p - 1 = p + 3 \\ \\ 2p - p = 3 + 1 \\ \\ p = 4$

So, p = 4 !!!!!!!!!!!

Answer 2

Hey !

Here is your answer !

____________

◾ If it is in AP

⭐ Then !

➡ b - a = c - b

▶ 2b = a + c


✴ 2 ( p + 3 ) = ( p - 1 ) + ( 3p - 1 )

✴ 2p + 6 = p - 1 + 3p - 1

✴ 2p + 6 = 4p - 2

✴ 8 = 2p

✴ p = 4

Hence answer is 4 !✔

Hope it helps !☺