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

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Solution

If the numbers a,b and c are in AP, then
b-a=c-b
$\Rightarrow 2 b = a + c$
The given three numbers (2p-1),(3p+1) and 11 are in AP.
Then,
$2 (3 p + 1) = (2 p - 1) + 11$
$\Rightarrow 6 p + 2 = 2 p - 1 + 11$
$\Rightarrow 6 p + 2 = 2 p + 10$
$\Rightarrow 6 p - 2 p = 10 - 2$
$\Rightarrow 4 p = 8$
$\Rightarrow p = 8 / 4$
$\Rightarrow p = 2$
So, the value of p is 2.
Then, the given numbers be
$2 \times 2 - 1, 3 \times 2 + 1, 11$
$4 - 1, 6 + 1, 11$
$3, 7, 11$

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

If the numbers a,b and c are in AP, thenb-a=c-b⇒2b=a+cThe given three numbers (2p-1),(3p+1) and 11 are in AP.Then,2(3p+1)=(2p−1)+11⇒6p+2=2p−1+11⇒6p+2=2p+10⇒6p−2p=10−2⇒4p=8⇒p=8/4⇒p=2So, the value of p is 2.Then, the given numbers be2×2−1,3×2+1,114−1,6+1,113,7,11