Question

$\bf \large\red{If \: {p}^{th} , \: {q}^{th} \: and \: {r}^{th} \: terms \: of \: an \: A.P \: are} \\ \bf \large \red{l \: , \: m \: , \: n \: respectively ,\: show \: that} \\ \sf \large \purple{(q - r)l \: + \: (r - p)m \: + \: (p - q)n \: = \: 0}$

Chapter :- Arithmetic Progression.​

Answer 1

Answer:

let A = fast term of the AP

and

let d= common difference of the AP

Now A=a+(p-1).d......(1)

B=A+(q-1).d......(2)

C=A+(r-1).d......(3)

Subtracting 2nd from 1st,3rd from 2ndand 1st from 3rd

we get

A-b=(p-q).d......(4)

b-c=(q-r).d........(5)

C-A=(r-p).d......(6)

multiply 4,5,6 by c,a, b respectively we have

C.(a-b)=c.(p-q).d.....(4).

a.(b-c)=a.(q-r).d.......(5)

b.(c-a)=b.(r-p).d.......(6)

A(q-r).d+b(r-p).d+c(p-q).d=0(a(q-r+b(r-p)+c(p-q).d=0

Now since d is common difference it should be non zero

.Hence,

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

(Ans)

Answer 2

Required Answer:-

Given:

• pth, qth and rth terms of an A.P. are l, m and n.

To prove:

• l(q-r) + m(r-p) + n(p-q) = 0

Proof:

We know that,

★ nth term of A.P. = a + (n - 1)d

So,

➡ l = a + (p - 1)d ...... (i)

➡ m = a + (q - 1)d ..... (ii)

➡ n = a + (r - 1)d ...... (iii)

So, Taking LHS,

l(q - r) + m(r - p) + n(p - q)

= (a + (p - 1)d)(q - r) + (a + (q - 1)d)(r - p) + (a + (r - 1)d)(p - q)

= a(q - r) + d(p - 1)(q - r) + a(r - p) + d(q - 1)(r - p) + a(p - q) + d(r - 1)(p - q)

= a[q - r + r - p + p - q] + d[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)]

= a × 0 + d[pq - pr - q + r + qr - pq - r + p + pr - rq - p + q]

= d[pq - pq + pr - pr + rq - rq + p - p + q - q + r - r]

= d[0 + 0 + 0 + 0 + 0 + 0]

= d × 0

= 0

= RHS (Hence Proved)

Learn More:

• A.P.:- Stands for Arithmetic Progression. It is a series where difference between two consecutive term is same (constant)