If the p^{th},q^{th},and r^{th} terms of an A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $A$ be the first term and $D$ be the common difference of the $A.P.$The $p^{th}$ term is $a = A + (p-1)D$ $...(1)$The $q^{th}$ term is $b = A

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Let $A$ be the first term and $D$ be the common difference of the $A.P.$The $p^{th}$ term is $a = A + (p-1)D$ $...(1)$The $q^{th}$ term is $b = A + (q-1)D$ $...(2)$The $r^{th}$ term is $c = A + (r-1)D$ $...(3)$Now,substitute these values into the expression $a(q-r) + b(r-p) + c(p-q)$:$= [A + (p-1)D](q-r) + [A + (q-1)D](r-p) + [A + (r-1)D](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 - qp - r + p + rp - rq - p + q]$$= 0 + D[0] = 0$Thus,the value of the expression is $0$.

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of an $A.P.$ are $a, b,$ and $c$ respectively,find the value of $a(q-r) + b(r-p) + c(p-q)$.

Similar Questions

The first term of an $A.P.$ of consecutive integers is $p^2 + 1$. The sum of $(2p + 1)$ terms of this series can be expressed as:

If $t_n = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1, 2, 3, \dots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \dots + \frac{1}{t_{2003}} = $

If $n$ is odd or even,then the sum of $n$ terms of the series $1 - 2 + 3 - 4 + 5 - 6 + \dots$ will be

The sum $\sum_{k=1}^{20}(1+2+3+\ldots+k)$ is

The $5^{th}$ term of a $G.P.$ is $2,$ find the product of first $9$ terms.

Vedclass Test Series

Exam Paper Generator

Online Exam Module