In arithmetic series:
an = a1 + ( n - 1 ) d
a1 = the initial term
d = the common difference of successive members
The sum of the first n terms:
Sn = n ( a1 + an ) / 2
In this case:
Sn = n [ a1 + a1 + ( n - 1 ) d ] / 2 =
n [ 2 a1 + ( n - 1 ) d ] / 2 =
[ 2 a1 + n • ( n - 1 ) d ] / 2 =
[ 2 a1 + n • d • ( n - 1 ) ] / 2 =
( 2 a1 n + n^2 d - n d ) / 2
Since:
2 n ( n + 3 ) = 2 n^2 + 6 n
( 2 a1 n + n^2 d - n d ) / 2 = 2 n^2 + 6 n
Multiply both dides by 2
2 a1 n + n^2 d - n d = 4 n^2 + 12 n
d n^2 + 2 a1 n - n d = 4 n^2 + 12 n
d n^2 + ( 2 a1- d ) n = 4 n^2 + 12 n
Comparing the left and right sides of the equation:
d n^2 = 4 n^2
d = 4
( 2 a1- d ) n = 12 n
2 a1- d = 12
2 a1- 4 = 12
Add 4 to both sides
2 a1= 16
a1 = 16 / 2
a1 = 8
the sum of the first n therms of an arithmetic series is 2n(n+3), what is the first term of the series
1 answer