Question

How many terms of ap-6 -11/2 -5 -9/2....Are needed give their sum zero?

Answer 1

25 terms are needed to get their sum zero.

Explanation:

Sum of first n terms in A.P. : $S_n=\dfrac{n}{2}(2a+(n-1)d)$ , where a= fisrt term , d= common difference.

Given A. P. = $-6 , \dfrac{-11}{2} , -5 , -\dfrac{-9}{2}$

Here , a= -6

$d=-\dfrac{11}{2}-(-6)=-\dfrac{11}{2}+6=\dfrac{-11+12}{2}=\dfrac{1}{2}$

To find , when $S_n=0$

$\dfrac{n}{2}(2(-6)+(n-1)(\dfrac{1}{2}))=0\\\\n(-12+\dfrac{n}{2}-\dfrac{1}{2})=0\\\\\text{either n=0 or } \dfrac{n}{2}-\dfrac{25}{2}=0\\\\ \text{either n=0 or } \dfrac{n}{2}=\dfrac{25}{2}\\\\ \text{either n=0 or } n=25$

Since n cannot be 0 , so n= 25

Hence, 25 terms are needed to get their sum zero.

# Learn more :

if the sum of 8 terms of an ap is 64 and the sum of 19 terms is 361. find the sum of n terms.

https://brainly.in/question/4385857

Answer 2

Answer:

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