Question

how many terms of the AP 18 ,15,12 are needed to give the sum 45

Answer 1

S = n/2(2a+(n-1)d)
45 = n/2 (36+(n-1)-3)
90= 36n-3n^2-3n
90=33n-3n^2
90-33n+3n^2=0
30-11n+n^2=0
30-6n-5n+n^2=0
6(5-n)-n(5-n)=0
n= 5 or 6
hope it will help

Answer 2

SN=n/2[2a+(n-1)d]
a=18
d=a2-a1=15-18=(-3)
SN=45

we need to find n

45=n/2[2(18)+(n-1)-3].........1
45=n/2[36+(-3n+3)]
90=n[36-3n+3]
90=n[39-3n]
90=39n-3n^2
3n^2-39n+90=0. ( divided by 3)
n^2-13n+30=0
n=10 r n=3 ( by factorisation method)

sub n in 1
45=10/2[2(18)+(10-1)-3]
45=5[36+(9)-3]
45=5[36-27]
45=5(9)
45=45

therefore n=10

I hope this will help you

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