Here are the first four terms of an arithmetic series K , 3k/4 , k/2 , k/4 Given that the 15th term of the series is ( 90 + 2k ) Calculate the sum of the first 30 terms

Answer 1

common difference = -k/4

a1=k

an=a1+d(n-1)

d=-k/4

an= k + (-k/4)(n-1)

a15=k+ (-k/4)(14)

=k-14k/4=k-7k/2=2k/2-7k/2

=(2k-7k)/2 =

-5k/2 = 90+2k

-5k=180+4k

9k=-180

k=-20=a1

d=-k/4=--20/4 =5

a15=90+2k=90-40=50

a15=-5k/2 = -5(-20)/2=50

a1 through a15 =

-20,-15,-10,-5,0,5,10

15,20,25,30,35,40,45,50

a15=50

sn = (n/2)(a1+an)

s30=(30/2)(-20+50)=15(30)=450

36,27,18,9,0,-9,-18,-27,-36,-45,-54,-63,-72,-81,-90

s15=-135x3=-405