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3 years ago

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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. . 2

09

2 answers:

Tems11 [23]3 years ago

8 0

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

Marina CMI [18]3 years ago

3 0

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

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