One of the basic question from arithmetic progression chapter in which we have been asked to find the number of terms required so that the sum must be 693 of the given series.
Book – RS Aggarwal, Class 10, chapter 5C, question no 10.
Rajan@2021Guru
How many terms of the AP 63, 60, 57, 54,….must be taken so that their sum is 693? Explain the double answer.
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Consider
a= First term
d= Common difference
n= number of terms
Here, a=63,d=60−63=−3 and sum =Sn=693
Sn=n/2[2a+(n−1)d]
693=n/2[2×63+(n−1)(−3)]
693×2=n(126−3n+3)
1386=n(129−3n)
1386=129n−3n2
3n^2−129n+1386=0
n^2−43n+462=0
Which is a quadratic equation
n^2−21n−22n+462=0
n(n−21)−22(n−21)=0
(n−21)(n−22)=0
Either, n−21=0, then n=21
or n−22=0, then n=22
Number of terms=21 or 22
T22=a+(n−1)d
=63+(22−1)(−3)
=63+21×(−3)=63−63=0
Which shows that, 22th term of AP is zero.
Number of terms are 21 or 22. So there will be no effect on the sum.