Question

Find the sum, S50, of the first 50 terms of the following arithmetic series:(−12)+(−9)+(−6

Answer 1

Answer:

S₅₀ = 3075

Step-by-step explanation:

Find the sum, S50, of the first 50 terms of the following arithmetic series:(−12)+(−9)+(−6)

an = a₁ + (n - 1) d

where:

a₁ = 1st term                      = -12

n = nth term                      = 50

d = common difference  

  = a₂-a₁ = -9 - (-12)            =3

plugin values into the formula:

an = -12 + (50 - 1) 3

    = 135

sum of an arithmetic sequence:

Sn = n (a1 + an)  

              2

S₅₀  = 50 ( -12 + 135 )

                    2

S₅₀ = 3075

Answer 2

Answer:

3075

Step-by-step explanation:

So we have the sequence:

$-12, -9, -6...$

As given, this is an arithmetic sequence and we want to find the sum of the first 50 terms.

The formula for the sum of an arithmetic sequence is:

$S=\frac{k}{2}(a+x_k}$

Where k is the number of terms, a is the initial term, and x_k is the last term.

We already know that k is 50 because we want to find the sum of the first 50 terms. a is -12 because it's the first number in our sequence. To find our last term, we need to do some more work.

To find the 50th term, let's write an explicit formula for our sequence. The standard form for the explicit formula for an arithmetic sequence is:

$x_n=a+d(n-1)$

Where a is the first term, d is the common difference, and n is the nth term.

We already know that a, the first term, is -12. From the sequence, we can see that the common difference d is +3 because each term is 3 more than the previous one. -12 plus 3 is -9, plus 3 is -6, and so on.

So, substitute -12 for a and 3 for d gives us:

$x_n=-12+3(n-1)$

So, to find the 50th term for our sum, substitute 50 for n:

$x_{50}=-12+3(50-1)$

Subtract:

$x_{50}=-12+3(49)$

Multiply:

$x_{50}=-12+147$

Add:

$x_{50}=135$

So, the last term of our 50 number sequence is 135.

Now, let's go back to our original formula:

$S=\frac{k}{2}(a+x_k})$

Now we know all the values. So, substitute 50 for k (the amount of terms), -12 for a (the first term), and 135 for k (the 50th and last term). So:

$S=\frac{50}{2}(-12+135)$

Divide:

$S=25(-12+135)$

Add:

$S=25(123)$

Multiply:

$S=3075$

So, our sum is 3075.

And we're done!