Question

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

Answer 1

A = 12, d = 5,  n = 11

The nth term: =    a + (n - 1)d =  12 + (11 - 1)*5 = 12 + 10*5 = 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

Answer 2

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