ICSE & CBSE Board Question Based on Arithmetic Progression of RS Aggarwal
Here you have to find the
(i) Number of terms so that their sum is Given
(ii)Explain the double answer
This is the Question Number 22 Of Exercise 11 C of RS Aggarwal Solution.
Deepak BoraNewbie
Find the number of terms of the AP 18,15,12,……so that their sum is 45. Explain the double answer.
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Given AP
18,15,12,….
First term of the AP (a)= 18
Common difference=-3
=a_{2}-a_{1}
=15-18
=-3
Let the sum on n term is 45
S_n=45
n/2(2a+(n-1)d)
45=n/2(2a+(n-1)d)
45=n/2[2(18)+(n-1)(-3)]
45=n/2[36+3n+3]
45×2=n[39+3n]
90=39n+3n^2
3n^2+39-90=0—————-divided by 3
n^2+13-30=0
n^2-10n-3n+30=0
n(n-10)+3(n-10)=0
(n-10)(n+3)=0
n=10
n=3