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sattari [20]
3 years ago
5

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
Mathematics
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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Collect like terms;

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5 0
3 years ago
How do I solve the problem
Grace [21]
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Solution 2: 3x - 2y = 5

Graphic Organizer: 2(y) + 5(x) = 3
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x = 2(y)÷ 3 + 5÷3

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Your answer would have to be

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~

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