Question

find tne number of term of the AP 18,15,12,....so that their sum is 45.explain the double answer.​

Answer 1

given that:

ap:18,15,12

first term a=18

common difference d=15-18=-3

we know that d=2nd term-1st term

sum of n terms(Sn=45)

formula:

Sn=n/2*(2a+(n-1)d)

Sn=n/2*(2×18+(n-1)-3)

45=n/2*(36-3n+3)

45×2=n*(36-3n+3)

90=n*(39-3n)

90=39n-3n^2

90=3(13-n^2)

90/3=(13-n^2)

30=(13-n^2)

n^2+(-13n)+30=0

now

30=-10*-3

so

n^2+(-3n-10n)+30=0

n(n-3)-10(n-3)=0

n-3=0

or

n-10=0

therefore the value is

n=10 or 3

hence,Ap is decreasing order in which it will include negative no's too in AP,so there is a possibility for the sum to be 45 for 10 terms and also for the first 3 terms.

I hope it helps you......