If p times the {p^{th}} term of an A.P. is eq | vedclass

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

Question

(a) $p\{ a + (p - 1)\,d\} = q\{ a + (q - 1)\,d\} $

$ \Rightarrow $ $a(p - q) + ({p^2} - {q^2})\,d + (q - p)\,d = 0$

$ \Rightarrow $ $(p - q)\{ a

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(a) $p\{ a + (p - 1)\,d\} = q\{ a + (q - 1)\,d\} $

$ \Rightarrow $ $a(p - q) + ({p^2} - {q^2})\,d + (q - p)\,d = 0$

$ \Rightarrow $ $(p - q)\{ a + (p + q - 1)\,d\} = 0$

$ \Rightarrow $ $a + (p + q - 1)\,d = 0$

$ \Rightarrow $${T_{p + q}} = 0$,

$\ \{ \because \;p 
e q\} $ .

If $p$ times the ${p^{th}}$ term of an $A.P.$ is equal to $q$ times the ${q^{th}}$ term of an $A.P.$, then ${(p + q)^{th}}$ term is

Similar Questions

If the variance of the terms in an increasing $A.P.$, $b _{1}, b _{2}, b _{3}, \ldots b _{11}$ is $90,$ then the common difference of this $A.P.$ is

If the first term of an $A.P.$ is $3$ and the sum of its first four terms is equal to one-fifth of the sum of the next four terms, then the sum of the first $20$ terms is equal to

The houses on one side of a road are numbered using consecutive even numbers. The sum of the numbers of all the houses in that row is $170$ . If there are at least $6$ houses in that row and $a$ is the number of the sixth house, then

Let the digits $a, b, c$ be in $A.P.$ Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in $A.P.$ at least once. How many such numbers can be formed?

The sum of the numbers between $100$ and $1000$, which is divisible by $9$ will be