If the sum of p terms of an arithmetic progre | vedclass

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

Question

(C) Given that the sum of $p$ terms equals the sum of $q$ terms,we have:$\frac{p}{2}[2a + (p - 1)d] = \frac{q}{2}[2a + (q - 1)d]$Expanding both sides:

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given that the sum of $p$ terms equals the sum of $q$ terms,we have:$\frac{p}{2}[2a + (p - 1)d] = \frac{q}{2}[2a + (q - 1)d]$Expanding both sides:$p[2a + (p - 1)d] = q[2a + (q - 1)d]$$2ap + p(p - 1)d = 2aq + q(q - 1)d$Rearranging the terms:$2a(p - q) + d[p(p - 1) - q(q - 1)] = 0$$2a(p - q) + d[p^2 - p - q^2 + q] = 0$$2a(p - q) + d[(p^2 - q^2) - (p - q)] = 0$Factoring out $(p - q)$:$(p - q)[2a + d(p + q - 1)] = 0$Since $p 
eq q$,we must have:$2a + (p + q - 1)d = 0$The sum of $(p + q)$ terms is given by:$S_{p+q} = \frac{p+q}{2}[2a + (p + q - 1)d]$Substituting the value $2a + (p + q - 1)d = 0$:$S_{p+q} = \frac{p+q}{2} 	imes 0 = 0$

If the sum of $p$ terms of an arithmetic progression is equal to the sum of its $q$ terms,then what is the sum of its $(p + q)$ terms?

Similar Questions

If the $p^{th}$ term of an arithmetic progression is $q$ and the $q^{th}$ term is $p$,then its $n^{th}$ term is:

If the sum of the first $p$ terms of an $A.P.$ is equal to the sum of the first $q$ terms,then find the sum of the first $(p+q)$ terms.

The number of terms of an $A.P.$ is even; the sum of all the odd terms is $24$,the sum of all the even terms is $30$ and the last term exceeds the first by $\frac{21}{2}$. Then the number of terms which are integers in the $A.P.$ is:

If $\log_{3} 2, \log_{3} (2^{x} - 5)$ and $\log_{3} (2^{x} - \frac{7}{2})$ are in an Arithmetic Progression $(AP)$,then $x = \dots$

Let $f(x)$ be a polynomial function of second degree. If $f(1) = f(-1)$ and $a, b, c$ are in $A.P.$,then $f'(a), f'(b)$ and $f'(c)$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module