if 8,p,q and 26 is an arithmetic progression find the values of p and q
a = 8
a + d = p ---> 8+d = p
a + 2d = q ---> 8 + 2d = q
a + 3d = 26
3d = 18
d = 6
then p = 8+6 = 14
q = 8+12 = 20
check:
so 8, 14, 20, 26 form an AS ? YES
To find the values of p and q in the arithmetic progression 8, p, q, 26, we need to use the formula for arithmetic progression:
an = a1 + (n-1)d,
where a1 is the first term, an is the nth term, and d is the common difference.
Given:
a1 = 8
a2 = p
a3 = q
a4 = 26
Using the formula, we can find the common difference (d) by subtracting the first term (a1) from the second term (a2):
d = a2 - a1
Substituting the values:
d = p - 8
Similarly, we can find the common difference (d) by subtracting the second term (a2) from the third term (a3):
d = a3 - a2
Substituting the values:
d = q - p
Since these two expressions both represent the common difference (d), we can equate them:
p - 8 = q - p
Adding p to both sides of the equation:
2p - 8 = q
Now, we can find the common difference (d) by subtracting the third term (a3) from the fourth term (a4):
d = a4 - a3
Substituting the values:
d = 26 - q
Since this expression also represents the common difference (d), we can equate it to the previous expression:
26 - q = 2p - 8
Simplifying the equation:
q - 2p = -18 (rearranging terms)
From these two equations:
2p - 8 = q (equation 1)
q - 2p = -18 (equation 2)
We can solve this system of equations by substitution or elimination method:
Using the elimination method:
Multiply equation 2 by 2 to get:
2q - 4p = -36
Now, add this equation to equation 1:
2p - 8 + 2q - 4p = q - 2p + (-36)
Simplifying:
-2p + q = -44
Rearranging the terms:
q - 2p = -44
Comparing it to equation 2, we see that the two equations are the same. This means that the system of equations has infinitely many solutions. Therefore, the values of p and q can be any numbers as long as they satisfy the equation q - 2p = -44.