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

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Answer 1 · 2017-07-08T06:27:53

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#

Answer 2 · 2017-07-08T06:35:10

#n=14#

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.