This question is from arithmetic progression chapter in which we have to find the number of terms of AP so that the sum must be 0 of the given this series.
Book – RS Aggarwal, Class 10, chapter 5C, question no 9.
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Consider
a= First term
d= Common difference
n= number of terms
Here,
a=9,d=17−9=8
Sum of terms =Sn=636
(n/2)[2a+(n−1)d]=636
(n/2)[2(9)+(n−1)(8)]=636
(n/2)[10+8n]=636
4n2+5n−636=0 (which is a quadratic equation)
(n−12)(4n+53)=0
Either (n−12)=0 or (4n+53)=0
n=12 or n=−53/4
Since n can’t be negative and fraction, so
n=12
Number of terms =12 terms.