What is the sum of the first 11 terms of the arithmetic series in which a1 = 12 and d = 5?. (Points : 1). 38. . 549. . 143. . 209
first we will find the 11th term
an = a1 + (n-1) d
a11 = 12 + (11 - 1) 5
a11 = 12 + 10 * 5
a11 = 12 + 50
a11 = 62
now we use the sum formula...
Sn = (n (a1 + an)) / 2
S11 = (11 (12 + 62)) / 2
S11 = (11 (74)) / 2
S11 = 814/2
S11 = 407
a = 12, d = 5, n = 11
The nth term: = a + (n - 1)d = 12 + (11 - 1)5 = 12 + 105 = 12 + 50 = 62
so the last term which is the eleventh, l = 62
Sum of first 11 = n/2(a + l), where l = last term.
S = 11/2(12 + 62)
= 11/2 74 = 11 37 = 407
Sum of the first 11 terms = 407