Question

How many terms of series -12-9-6-3....must be taken so that the sum may be 54

Answer 1

hi friend,

given series is in arithmetic progression

with a=-12 and d=3

sum of n terms of an AP=n/2(2a+(n-1)d)

=>54=n/2(2(-12)+(n-1)3)

=>54=n/2(-24+3n-3)

=>108=n(3n-27)

dividing with 3 on boths sides, we get

=>36=n²-9n

=>n²-9n-36=0

=>n²-12n+3n-36=0

=>n(n-12)+3(n-12)=0

=>(n+3)(n-12)=0

n=12

I hope this will help u :)

Answer 2

This is arithmetic progression ,with starting term a=-12,d=+3

⇒ Sum of 'n ' terms in a A.P =n/2(2a+(n-1)d

⇒54=n/2(2*-12+(n-1)3)

⇒54=n/2 (-24+3n-3)

⇒54=n/2(-27+3n)

⇒108=n(3n-27)

⇒3n²-27n-108=0

⇒3(n²-9n-36)=0

⇒n²-9n-36=0

⇒n²-12n+3n-36=0

⇒n(n-12)+3(n-12)=0

⇒(n-12)(n+3)=0

⇒n= 12 or -3

Number of terms cant be negative

so,

Number of terms =12

hope helped!!