Question

How many terms of the series of - 3+0+3+6+9+...are needed to give a sum of 105?​

Answer 1

Answer:

10

Step-by-step explanation:

Remember that the formula for the sum of an arithmetic series 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 of the series.

We essentially want to find k, the number of terms, given that the sum S is equal to 105. So, substitute 105 into our equation:

$105=\frac{k}{2}(a+x_k)$

To do so, we need to final term x_k. We don't know what it is yet, but that doesn't matter. All we need to do is to write it in terms of k. First, remember that the standard form for the explicit formula of 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.

From our sequence, we can see that the first term is -3.

Also, we can determine that our common difference is +3, since each subsequent term is 3 more than the previous one. -3+3 is 0, 0+3 is 3, 3+3 is 6, and so on.

Therefore, our explicit formula is:

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

Therefore, our final term, x_k, will be if we substitute k for n. So, we can acquire the equation:

$x_k=-3+3(k-1)$

Now that we know what x_k is, we can substitute that into our original equation:

$105=\frac{k}{2}(a+x_k)$

Substitute the equation into x_k. Also, let's substitute -3 (our first term) for a. So:

$105=\frac{k}{2}(-3+(-3+3(k-1)))$

And now, all we have to do is to solve for k.

First, distribute the 3:

$105=\frac{k}{2}(-3+(-3+3k-3))$

Add within the parentheses:

$105=\frac{k}{2}(3k-9)$

Multiply both sides by 2. This removes the fraction on the right:

$210=k(3k-9)$

Distribute. We will get a quadratic:

$210=3k^2-9k$

So, let's solve for k. Let's divide everything by 3:

$70=k^2-3k$

Subtract 70 from both sides:

$0=k^2-3k-70$

Factor. We can use -10 and 7. So:

$0=(k-10)(k+7)$

Zero Product Property:

$k-10=0\text{ or } k+7=0$

Solve for k for each equation:

$k=10\text{ or } k=-7$

-7 doesn't make sense (we can't have -7 terms). Remove that solution. So, we are left with:

$k=10$

Therefore, the number of terms we have in our series for our sum to be 105 is 10.

And we're done!