Okay, so I know this should be really easy but my mind is just drawing a blank:
For each series, calculate t12 and S12
a) 37 + 41 + 45 + 49 + ...
Okay so I calculated S12 to be 708, and I know that's right.
For t12, I know I can easily find the general formula and then use that to find t12, but isn't there a faster way to do it knowing that the sum is 708?
Oh. I guess the point of the question was for you to find t12 first by using the general formula so that you could have the last term, which would allow us to use the simpler arithmetic series formula, i.e., n(t1+tn)/2 rather than n/2(2a+(n-1)d)
But I'd still like to know if there IS a way to find t12 with S12, just in case that comes up later or something.
Yes, there is a faster way to find t12 for this arithmetic series without explicitly finding the general formula. Since you already know the sum of the first 12 terms is 708 (S12 = 708), you can use the formula for the sum of an arithmetic series to find the average of all the terms.
The formula for the sum of an arithmetic series is:
Sn = (n/2)(2a + (n-1)d)
Where:
Sn refers to the sum of the first n terms,
a is the first term of the series,
d is the common difference between consecutive terms, and
n is the number of terms in the series.
In your case, the sum of the first 12 terms (S12) is given as 708.
We can rearrange the formula to solve for the first term (a) in terms of the sum (S), the common difference (d), and the number of terms (n):
2a + (n-1)d = 2S/n
Now, substitute the known values into the equation:
2a + 11d = 2(708)/12
2a + 11d = 118
Since the series is arithmetic, the difference between consecutive terms is constant. Therefore, you can express any term of the series in terms of the first term and the common difference as follows:
tn = a + (n-1)d
Now, substitute n = 12 into the equation and the known value of the sum S12 = 708:
t12 = a + 11d
At this point, we have a system of equations:
2a + 11d = 118
t12 = a + 11d
To solve this system, you can use substitution or elimination. From the first equation, you can solve for a in terms of d:
2a = 118 - 11d
a = (118 - 11d)/2
Now, substitute this value of a into the second equation:
t12 = (118 - 11d)/2 + 11d
t12 = 118/2 + 11d/2
t12 = 59 + 11d/2
So, using the given sum S12 = 708, the twelfth term t12 is represented by the equation t12 = 59 + 11d/2.