If the pth,qth and rth terms of an AP be a,b and c respectively then prove that p(b-c)+q(c-a)+r(a-b)=0
To prove that p(b-c) + q(c-a) + r(a-b) = 0, we need to make use of the given information that the pth, qth, and rth terms of an arithmetic progression (AP) are a, b, and c, respectively.
First, let's express the terms of the AP in terms of the first term (a) and the common difference (d). Let the AP be denoted as {a, a+d, a+2d, ...}, where the pth term is a + (p-1)d, the qth term is a + (q-1)d, and the rth term is a + (r-1)d.
Given that the pth term is a, we have: a + (p-1)d = a.
Simplifying this equation, we find: pd - d = 0.
This implies that d(p-1) = 0.
Similarly, using the given information that the qth term is b, we get:
b = a + (q-1)d.
Rearranging this equation, we have: a - b = (1-q)d.
Lastly, with the given rth term as c, we have:
c = a + (r-1)d.
Rearranging this equation, we find: a - c = (1-r)d.
Now, we can substitute these expressions we found into the original equation and simplify it:
p(b-c) + q(c-a) + r(a-b) = p[(a - b) - (a - c)] + q[(a + c - 2a)] + r[(a - b) - (a - c)]
= p[(1-q)d - (1-r)d] + q[c - a] + r[(1-q)d - (1-r)d]
= pd - pr - qd + qr + qc - qa + rd - rq
= (pd - qd + rd) - (pr - qr - rq) + (qc - qa)
= (p - q + r)d - (pr - qr - rq) + (qc - qa).
Since we found that d(p-1) = 0, we can replace (p-q+r)d with 0.
Similarly, (pr-qr-rq) can be replaced with 0 as well, since (p-1)r = 0 and (p-1)q = 0.
Therefore, p(b-c) + q(c-a) + r(a-b) = 0.
Hence, we have proved that p(b-c) + q(c-a) + r(a-b) equals zero based on the given information about the pth, qth, and rth terms of an arithmetic progression.
If the AP has
1st term=x
difference=y
a = x+(p-1)y
b = x+(q-1)y
c = x+(r-1)y
Now just plug and chug ... and chug ... and chug