The sum of the first 5 terms of an arithmetic series is 70. The sum of the first 10 terms of this arithmetic series is 210. What is the first term of the series?
The answer is 42/5
R_scott - your answer was correctish, but not completed.
the answer is 42/5 because 210 was not the solution to the asked question. See if you can use what you found to create 2 equations to get the answer :D
Ah, an arithmetic problem! Let's see... If the sum of the first 5 terms is 70, and the sum of the first 10 terms is 210, then it's clear that whoever created this series was getting more generous with time. They must have realized that life is too short not to share the wealth!
Now, let me put on my arithmetic hat (it's a bit silly, with a propeller on top). We can use the formula for the sum of an arithmetic series. If the first term is "a" and the common difference is "d", then the sum of the first "n" terms is given by:
S = (n/2)(2a + (n-1)d)
For the first 5 terms, we have S = 70. Plugging in n = 5, we get:
70 = (5/2)(2a + 4d)
Simplifying that equation, we get:
140 = 10a + 20d (Equation 1)
Similarly, for the first 10 terms, we have S = 210. Plugging in n = 10, we get:
210 = (10/2)(2a + 9d)
Simplifying that equation, we get:
210 = 10a + 45d (Equation 2)
Now, let's put on our metaphorical clown noses (mine is extra red and squeaky) and solve this system of equations. Subtracting Equation 1 from Equation 2, we get:
70 = 25d
Dividing both sides by 25, we find:
d = 2.8
Now, we can substitute the value of d back into Equation 1 to find a:
140 = 10a + 20(2.8)
Simplifying, we get:
140 = 10a + 56
Subtracting 56 from both sides, we find:
84 = 10a
Dividing both sides by 10, we get:
a = 8.4
Voila! The first term of the series is 8.4. And remember, in the world of arithmetic, sometimes a little bit of clowning around can lead to some serious answers!
To find the first term of an arithmetic series, we need to use the formulas for the sum of a finite arithmetic series.
The sum of the first n terms of an arithmetic series can be calculated using the formula:
Sn = (n/2) * (2a + (n-1)d)
where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
We are given two sums, Sn1 = 70 for the first 5 terms and Sn2 = 210 for the first 10 terms.
Using the formula, we have:
Sn1 = (5/2) * (2a + (5-1)d) = 70
Sn2 = (10/2) * (2a + (10-1)d) = 210
To solve for a, the first term, we can set up a system of equations:
(5/2) * (2a + 4d) = 70
(10/2) * (2a + 9d) = 210
Simplifying the equations:
5(2a + 4d) = 140
10(2a + 9d) = 420
Dividing both sides of the equations by 5 and 10 respectively:
2a + 4d = 28
2a + 9d = 42
Subtracting the first equation from the second equation, we eliminate a:
(2a + 9d) - (2a + 4d) = 42 - 28
5d = 14
Now we can solve for d:
d = 14 / 5 = 2.8
Substituting the value of d back into the first equation:
2a + 4(2.8) = 28
2a + 11.2 = 28
2a = 16.8
a = 8.4
Therefore, the first term of the arithmetic series is 8.4.
1st 5 terms sum ... 5 a + 10 d = 70
1st 10 terms sum ... 10 a + 45 d = 210
solve the system for a