The 11^{text{th}} term of the AP: -5, frac{-5 | vedclass

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

Question

(B) Given $AP$ is $-5, \frac{-5}{2}, 0, \frac{5}{2}, \dots$Here,the first term $a = -5$.The common difference $d = \frac{-5}{2} - (-5) = \frac{-5}{2}

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) Given $AP$ is $-5, \frac{-5}{2}, 0, \frac{5}{2}, \dots$Here,the first term $a = -5$.The common difference $d = \frac{-5}{2} - (-5) = \frac{-5}{2} + 5 = \frac{-5 + 10}{2} = \frac{5}{2}$.The formula for the $n^{	ext{th}}$ term of an $AP$ is $a_n = a + (n - 1)d$.To find the $11^{	ext{th}}$ term $(n = 11)$:$a_{11} = -5 + (11 - 1) 	imes \frac{5}{2}$$a_{11} = -5 + 10 	imes \frac{5}{2}$$a_{11} = -5 + 5 	imes 5$$a_{11} = -5 + 25$$a_{11} = 20$.

The $11^{\text{th}}$ term of the $AP: -5, \frac{-5}{2}, 0, \frac{5}{2}, \dots$ is

Similar Questions

The sum of the first $n$ even natural numbers is........

Find five numbers in $A.P.$ such that their sum is $30$ and the sum of their squares is $220$.

$A$ class has $49$ students who are allotted roll numbers from $1$ to $49$. Find a roll number $n$ such that the sum of all the roll numbers smaller than $n$ is equal to the sum of all the roll numbers greater than $n$.

The first term of an $A.P.$ is denoted by $\ldots \ldots \ldots \ldots .$

The sum of all the terms of a finite $A.P.$ having $n$ terms is given by $S_{n} = \ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module