Question

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

Answer:

The given three numbers (2p-1),(3p+1) and 11 are in AP. So, the value of p is 2.

Step-by-step explanation:

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Answer 2

If the numbers a,b and c are in AP, then

If the numbers a,b and c are in AP, thenb-a=c-b

If the numbers a,b and c are in AP, thenb-a=c-b⇒2b=a+c

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.

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,

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

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

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

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

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

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

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=2

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.

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 be

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+1,11

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+1,11

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