How to find the number of terms n in 1+(-2)+(-5)+(-8)...., S"n"=-259?

2 Answers
P dilip_k
Jul 8, 2017

The given series belongs to AP.

Its first term a=1 and

common difference d=-2-1=-3

So sum of n terms S_n=n/2(2xxa+(n-1)xxd)

Hence

n/2(2xxa+(n-1)xxd)=-259

=>n/2(2xx1+(n-1)xx(-3))=-259

=>2n-3n^2+3n=-259xx2

=>3n^2-5n-518=0

=>n=(5+sqrt((-5)^2-4xx3xx(-518)))/(2xx3)

=>n=(5+sqrt6241)/6

=>n=(5+79)/6=14

IDKwhatName
Jul 8, 2017

n=14

Explanation:

The sum of a serues is given by Sn=n/2[2a+d(n-1)]

You know that d=-2-1=-3, and that Sn=-259=n/2[2(1)+(-3)(n-1)]=n/2[2-3(n-1)]

To remove the 1/2, we multiply both sides by 2 to get -518=n[2-3n+3]=n[5-3n]=5n-3n^2

-518=5n-3n^2

Putting all terms to one side gives 3n^2-5n-518=0

We now that the only two factors of 3 are 1 and 3, so our quadratic is in the form of (3x+a)(x+b).

-518=-2*7*37=-14x37

a and b = -14 and 37

3(-14)+37=-5

So, (3n+37)(n-14)=0

As n cannot be negative or a decimal, n-14=0, n=14

The sum of the first 14 terms gives -259.

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