determine the 1st term of the first 21 term of an arithmetic series if the sum of the three middle term is 126 and the sum of the three last terms is 288

- 18

Step-by-step explanation:

arithmetic  sequences are in form:

a(n)=a(o)+d(n-1)

sum are in form:

s(n)=[a(0)+a(n)](n/2)

three middle terms is:

10th, 11th, 12th term

How to think: 21 = 10+1+10 --> middle 3 is 10,11,12

sum of these 3 terms:

a(10)+a(11)+a(12)

=a(o)+9d+a(0)+10d+a(0)+11d

=3a(0)+30d=126

a(0)+10d=42 (equation 1)

Last three terms is:

19th, 20th, 21th

a(19)+a(20)+a(21)

=a(0)+18d+a(0)+19d+a(0)+20d

=3a(0)+57d=288

a(0)+19d=96 (equation 2)

now you have

a(0)+10d=42 (equation 1)

a(0)+19d=96 (equation 2)

solve and gives

d=6

a(0)= -18

my acc plz

step-by-step explanation:

about 0.75 hours

step-by-step explanation: