And the Sum of the first 15 terms for:
2,4,6,8,10
an = a1 + (n - 1)d
d = Δ between consecutive terms
d = an - an - 1
We see a common difference = 2
We have a1 = 2
an = 2 + 2(n - 1)
Plug in n = 6 and d = 2
a6 = 2 + 2(6 - 1)
a6 = 12
a6 = 2 + 2(5)
a6 = 2 + 10
Plug in n = 7 and d = 2
a7 = 2 + 2(7 - 1)
a7 = 14
a7 = 2 + 2(6)
a7 = 2 + 12
Plug in n = 8 and d = 2
a8 = 2 + 2(8 - 1)
a8 = 16
a8 = 2 + 2(7)
a8 = 2 + 14
Plug in n = 9 and d = 2
a9 = 2 + 2(9 - 1)
a9 = 18
a9 = 2 + 2(8)
a9 = 2 + 16
Plug in n = 10 and d = 2
a10 = 2 + 2(10 - 1)
a10 = 20
a10 = 2 + 2(9)
a10 = 2 + 18
Plug in n = 11 and d = 2
a11 = 2 + 2(11 - 1)
a11 = 22
a11 = 2 + 2(10)
a11 = 2 + 20
Plug in n = 12 and d = 2
a12 = 2 + 2(12 - 1)
a12 = 24
a12 = 2 + 2(11)
a12 = 2 + 22
Plug in n = 13 and d = 2
a13 = 2 + 2(13 - 1)
a13 = 26
a13 = 2 + 2(12)
a13 = 2 + 24
Plug in n = 14 and d = 2
a14 = 2 + 2(14 - 1)
a14 = 28
a14 = 2 + 2(13)
a14 = 2 + 26
Plug in n = 15 and d = 2
a15 = 2 + 2(15 - 1)
a15 = 30
a15 = 2 + 2(14)
a15 = 2 + 28
Sn = | n(a1 + an) |
2 |
S15 = | 15(a1 + a15) |
2 |
S15 = | 15(2 + 30) |
2 |
S15 = | 15(32) |
2 |
S15 = | 480 |
2 |