Question

[CBSE 2008] [Ans
38. How many terms of the A.P. 3, 6, 9, 12, .... should be taken so tha
sum is 165?
An​

Answer 1

Answer:

a similar answer

Step-by-step explanation:

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