What is the sum of the arithmetic sequence 3, 9, 15..., if there are 26 terms?

2 answers:

7 0

9 - 3 = 6, 15-9 = 6 the difference is 6, So d = 6

First term: a1 = 3

Sn = n*(a1 + an)/2

Sn = n*(a1 + a1 + (n-1)*d)/2

Sn = n*(2*a1 + (n-1)*d)/2

substitute 26 for n
 S26 = 26*(2*a1 + (26-1)*d)/2

substitute 3 for a1

S26 = 26*(2*3 + (26-1)*d)/2

substitute 6 for d

S26 = 26*(2*3 + (26-1)*6)/2 

S26 = 2,028

marissa [1.9K]4 years ago

3 0

Sum of an arithmetic series is:

S = n*(a_1+a_n)/2, that easy.

n is the number of terms, 26 in your case. a_n is a_26. We need to find the series!

a_n = a_1 + (n-1)*d, the difference, d, is 6.

So a_n = 3 + (n-1)*6 = 6*n-3 (check it works: 3, 9, 15, ...)

Now a_26 = 6*26-3 = 153, so finally:

S = 26*(3+153)/2 = 26*78 = 2028

2028 is the answer

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If 30 buses can carry 1,500 people, how many people can 5 buses carry? A 200 © 500 B 250 D 750

maksim [4K]

Answer:

$\boxed {\boxed {\sf 250 \ people}}$

Step-by-step explanation:

Let's set up a proportion using the following setup.

$\frac {buses}{people}=\frac {buses}{people}$

We know 30 buses can carry 1,500 people.

$\frac {30 \ buses}{1500 \ people}=\frac {buses}{people}$

We don't know how many people 5 buses can carry, so we say 5 buses carry x people.

$\frac {30 \ buses}{1500 \ people}=\frac {5 \ buses}{x \ people}$

$\frac {30 }{1500 }=\frac {5 }{x }$

Cross multiply. Multiply the numerator of the first fraction by the second fraction's denominator. Then, multiply the first denominator by the second numerator.

$30*x=1500*5\\30x=7500$