Question

If the `p^(th), q^(th) and r^(th)` terms of an arithmetic progression are `P, Q, R` respectively, then `P(q-r) + Q(r-p) + R(p-q)` is equal to?

Answer 1

`P(q-r)+Q(r-p)+R(p-q)=0`

Explanation:

The `n^"th"` term `a_n` of an arithmetic progression with common difference `d` and initial term `a_1` is given by

`a_n = a_1+(n-1)d`

thus, we can write

`P = a_1+(p-1)d`
`Q = a_1+(q-1)d`
`R = a_1+(r-1)d`

If we subtract `R` from `Q`, we find

`Q-R = a_1+(q-1)d - (a_1+(r-1)d) = (q-r)d`

Thus `q-r = (Q-R)/d`

Similarly, we have `r-p = (R-P)/d` and `p-q = (P-Q)/d`

Substituting these into the given expression, we get

`P(q-r)+Q(r-p)+R(p-q)`

`=(P(Q-R))/d+(Q(R-P))/d+(R(P-Q))/d`

`=(PQ-PR + QR-PQ + PR-QR)/d`

`=(PQ-PQ+PR-PR+QR-QR)/d`

`=0/d`

`=0`