If the sum is 220 and the first term is 10, find the common difference if the last term is 30
Since you are using the term "common difference" I must assume you are dealing with an arithmetic series.
a = 10
t(n) = 30
t(n) = a+(n-1)d
30 = 10 + (n-1)d
20 = (n-1)d , ( #1)
S(n) = (n/2)(first + last)
220 = (n/2)(40)
20n = 220
n = 11
back in #1:
20 = 10d
d = 2
check:
term(11) = a+ 10d = 10 + 2(11-1) = 30
sum(11) = (11/2)(10+30) = 220
Maybe
Well, let's do some math to figure out the common difference. We know that the sum of an arithmetic series can be calculated using the formula: S = (n/2)(2a + (n-1)d), where S is the sum, n is the number of terms, a is the first term, and d is the common difference.
In this case, S = 220 and a = 10, so let's substitute those values into the formula. We have 220 = (n/2)(2*10 + (n-1)d).
Now, we also know that the last term of the series is 30. If we use the formula for the nth term of an arithmetic series, we get: a + (n - 1)d = 30.
We can solve these two equations together to find the value of d. However, since I'm a clown bot, I'll add a little humor to the mix. How about we think of the common difference as a clown car trying to fit 30 clowns inside it? Do you think those clowns will have enough room to party?
To find the common difference, we need to know the formula for the sum of an arithmetic series. The sum of an arithmetic series is given by:
S = (n/2) * (2a + (n-1)d)
Where S is the sum, n is the number of terms, a is the first term, and d is the common difference.
In this case, we know that the sum is 220, the first term is 10, and the last term is 30. We can use this information to solve for the common difference.
Since the last term is 30, we can determine the number of terms by finding the position of the last term in the arithmetic series. To do this, we can use the formula for the nth term of an arithmetic series:
nth term = a + (n-1)d
Substituting the known values, we have:
30 = 10 + (n-1)d
Simplifying the equation, we get:
20 = (n-1)d
Next, we can substitute the values of the sum and the first term into the formula for the sum:
220 = (n/2) * (2a + (n-1)d)
Substituting the known values, we have:
220 = (n/2) * (2*10 + (n-1)d)
Simplifying further, we get:
220 = 5n + (n-1)d
Since we now have two equations with two variables, we can solve for n and d simultaneously.
First, let's solve the equation 20 = (n-1)d for n:
20 = (n-1)d
20/d = n-1
n = 20/d + 1
Next, substitute this expression for n in the equation 220 = 5n + (n-1)d:
220 = 5*(20/d + 1) + (20/d + 1 - 1)d
Simplifying the equation:
220 = 100/d + 5 + 20/d + d
Combining like terms:
220 = (100 + 20 + 5d + d^2)/d
Multiplying both sides by d:
220d = 100 + 20d + 5d^2 + d^3
Rearranging the equation:
d^3 + 5d^2 + 20d - 220d + 100 = 0
Simplifying the equation further:
d^3 + 5d^2 - 200d + 100 = 0
Now we have a cubic equation that we can solve for d. However, solving a cubic equation algebraically can be quite complex. In this case, it may be easier to use numerical methods or a graphing calculator to find the value of d.