# An arithmetic series has first term -50 and common difference 4. How many terms are in the series so that the sum of the series first exceeds 100?

## Again, use your definition.

you know S(n) = (n/2)(2a + (n-1)d ), so

(n/2)(2a + (n-1)d ) > 100

and you know both a and d

remember that n has to be a whole number, and you will end up with a quadratic inequality.

(n/2)(2a + (n-1)d ) = 100 ,

sub in your values, solve the resulting quadratic for n

**round up** your n to the next whole number

## Oh, I love arithmetic! Let's calculate it step by step like a well-behaved calculating clown.

To find the number of terms, we need to determine when the sum of the series first exceeds 100.

The sum of an arithmetic series can be calculated using the formula: Sn = (n/2)(2a + (n-1)d), where n is the number of terms, a is the first term, and d is the common difference.

So, plugging in the given values:

100 = (n/2)(2 * -50 + (n-1) * 4)

We can simplify this equation and solve for n, just like a mathemagician:

100 = (n/2)(-100 + 4n - 4)

Now, let me do some clown calculations...

100 = (n/2)(-104 + 4n)

200 = n(-104 + 4n)

200 = -104n + 4n^2

Rearrange it to get:

4n^2 - 104n + 200 = 0

Now, let's solve it using the quadratic formula:

n = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values: a = 4, b = -104, and c = 200

n = (-(-104) ± √((-104)^2 - 4 * 4 * 200)) / (2 * 4)

n = (104 ± √(10816 - 3200)) / 8

Now, let's simplify:

n = (104 ± √(7626)) / 8

Unfortunately, the clown calculator in my head is not equipped to find the square root of 7626. So we may have reached the limits of clown math here.

But don't worry, my friend! You can use a calculator to find the approximate value of n. I apologize if I couldn't provide the exact answer this time, but I hope the process itself brought a smile to your face!

## To find the number of terms in the arithmetic series, we first need to find the formula for the sum of an arithmetic series.

The formula for the sum of an arithmetic series is:

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

Where:

Sn represents the sum of the series

n represents the number of terms

a represents the first term

d represents the common difference

In this problem, a = -50 and d = 4. We want to find the smallest number of terms (n) that makes the sum (Sn) exceed 100. Therefore, we can set up the following inequality:

Sn > 100

Substituting the values for a and d into the equation:

( n/2 ) * ( 2 * -50 + ( n - 1 ) * 4 ) > 100

Simplifying the equation:

( n/2 ) * ( -100 + 4n - 4 ) > 100

( n/2 ) * ( 4n - 104 ) > 100

Dividing both sides of the inequality by 2:

( 4n - 104 ) * n > 200

4n^2 - 104n > 200

Rearranging the inequality:

4n^2 - 104n - 200 > 0

Simplifying the equation:

n^2 - 26n - 50 > 0

To solve this inequality, we can find the range of values for n that will make it true.

## To find the number of terms in the arithmetic series, we need to determine when the sum of the series first exceeds 100.

The formula to find the sum of an arithmetic series is given by:

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

where:

Sn is the sum of the first n terms,

a is the first term of the series,

d is the common difference, and

n is the number of terms.

In this case, the first term (a) is -50, and the common difference (d) is 4. We want to find the value of n such that Sn exceeds 100, so we can set up the following inequality:

Sn > 100.

Replacing the values of a and d, we have:

(n/2) * (2(-50) + (n-1)(4)) > 100.

Simplifying the equation further:

(-100 + 4(n-1))n > 200.

Expanding and simplifying again:

-100n + 4n^2 - 4n > 200.

Rearranging the equation:

4n^2 - 4n - 100n > 200,

4n^2 - 104n - 200 > 0.

Now we have a quadratic inequality. To solve this, we can factorize, use the quadratic formula, or consider the factors of the quadratic equation.

Considering the factors, we can rewrite the equation as:

4(n - 10)(n + 5) > 0.

To determine when the inequality is true, we need to find the range of n. Since 4 is a positive number, we need both (n - 10) and (n + 5) to have the same signs.

Case 1: (n - 10) > 0 and (n + 5) > 0.

In this case, n is greater than 10 and greater than -5. Therefore, n > 10.

Case 2: (n - 10) < 0 and (n + 5) < 0.

In this case, n is less than 10 and less than -5. Therefore, n < -5.

So, the range of n for which the inequality is true is -5 < n < 10.

However, we are interested in the positive number of terms, so we can conclude that there are 9 terms in the series so that the sum of the series first exceeds 100.