Question

What is the sum of the finite arithmetic series 4+8+12+16...+76

Answer 1

A(n)=4+4(n-1)

a(n)=4+4n-4

a(n)=4n

76=4n

n=19

The sum of any arithmetic sequence (series are infinite) is:

(a+a(n))(n/2)

The average of the first and last terms times the number of terms, in this case we found that n=19 so:

19(4+76)/2=760

Answer 2

Answer:

The nth term of the arithmetic sequence is given by:

$l= a+(n-1)d$            .....[1]

and

the sum of the arithmetic sequence is:

$S_n = \frac{n}{2}(a+l)$              .....[2]

where,

a is the first term

d is the common difference of two consecutive term

l is the last term in the series

As per the statement:

The finite arithmetic series 4+8+12+16...+76

here, a = 4 and

Common difference(d) = 4

Since,

8-4 = 4,

12-8 = 4,

16-12 = 4 and so on

last term of the finite series(l) = 76

Substitute these in [1] we have

$76 = 4+(n-1)4$

Using distributive property, $a \cdot (b+c) =a\cdot b+ a\cdot c$

$76 = 4+4n -4$

Simplify:

$76 = 4n$

Divide both sides by 4 we have;

19 = n

or

n= 19

Substitute the given value and n = 19 in [2]

$S_{19} = \frac{19}{2}(4+76) = \frac{19}{2} \cdot 80 = 19 \cdot 40 = 760$

Therefore, the sum of the given finite arithmetic series is, 760