The pth, qth and rth terms of an A.P. are a, b, c respectively. Show that (q – r) a + (r – p) b + (p -q) c = 0

Asked by Abhisek | 1 year ago |  62

Solution :-

Let’s assume t and d to be the first term and the common difference of the A.P. respectively.

Then the nth term of the A.P. is given by, a= t + (n – 1) d

Thus,

ap = t + (p – 1) d = a  … (1)

aq = t + (q – 1) d = b  … (2)

ar = t + (r – 1) d = c  … (3)

On subtracting equation (2) from (1), we get

(p – 1 – q + 1) d = a – b

(p – q) d = a – b

$$d=\dfrac{a-b}{p-q}$$...........(4)

On subtracting equation (3) from (2), we get

(q – 1 – r + 1) d = b – c

(q – r) d = b – c

$$d=\dfrac{b-c}{q-r}$$.............(5)

Equating both the values of d obtained in (4) and (5), we get

$$\dfrac{a-b}{p-q}=\dfrac{b-c}{q-r}$$

$$(a - b)(q - r) = (b - c)(p - q)$$

$$aq - bq - ar+br =bp-bq-cp+cq$$

$$bp-cp+cq-aq+ar-br=0$$

By rearranging terms we get

$$(-aq+ar)+(bp-br)+(-cp+cq)=0$$

$$-a(q-r)-b(r-p)-c(p-q)=0$$

$$-a(q-r)+b(r-p)+c(p-q)=0$$

Therefore,$$(q-r)a+(r-p)b+(p-q)c=0$$ is proved.

Answered by Pragya Singh | 1 year ago

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