This is the question of arithmetic progression chapter in which we have been asked to find the number of terms must be taken so that their sum is 300 of the given series.
Book RS Aggarwal, Class 10, chapter 5C, question no 11.
Rajan@2021Guru
How many terms of the AP 20, 19 1/3, 18 2/3,….must be taken so that their sum is 300? Explain the double answer.
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Consider
a= First term
d= Common difference
n= number of terms
Here, a=20,d=−2/3 and sum =Sn=300(for n number of terms)
Sn=n/2[2a+(n−1)d]
300=n/2[2×20+(n−1)(−2/3)]
600=n[40−2n/3+2/3]
600=n[122/3−2n/3]
600=[n(122−2n)]/3
1800=122n−2n^2
2n^2−122n+1800=0
n^2−61n+900=0
n^2−25n−36n+900=0
n(n−25)−36(n−25)=0
(n−25)(n−36)=0
Either (n−25)=0 or (n−36)=0
n=25 or n=36
For n=25
300=25/2[2×20+(25−1)(−2/3)]
=25/2[40+24×(−2/3)]
=25/2[40−16]
=25/2×24
=300
Which is true.
For n=36
300=36/2[2×20+(36−1)(−2/3)]
=18[40+35×(−2/3)]
=18[40−70/3]
=18×(120−70)/3
=300
Which is true.
Result is true for both the values of n.
So both the numbers are correct.
Therefore, sum of 11 terms is zero. (36−25=11).