Expert: Ahmed Salami - 10/12/2008

Question
Hi, I am stuck with the following questions. I think you need to use simultaneous equations but I'm not sure how.

1. The sum of the first two terms of an arithmetic series is 47, and the thirtieth term of this series is -62. Find

a) The first term of the series and common difference
b) The sum of the first 60 terms of the series


2. The first term of an arithmetic series is -13 and the last is 99. The sum of the series is 1419. Find the number of terms and the common difference. Find also the sum of all the positive terms in the series.


thank you very much for your help!

Answer
Hi Sally,
Yes, you need to use simultaneous equation. Lets find out why and how!
1). For an arithmetic series the first term a, common difference d and nth term T(n) are related by the formula
T(n) = a + (n-1)d
And so the first term, second term and thirtieth term would be a, a + d, and a + 29d respectively
From the available information,
T(1) + T(2) = 47
and  T(30) = -62
Therefore
a + (a + d) = 47
2a + d = 47
and
a + 29d = -62
We have to solve the last two equations simultaneously. We first multiply the second equation by 2 to get
2a + 58d = -124
and then substract the first from it.
2a + 58d = -124
-
2a + d = 47
We end up with
57d = -171
d = -171/57 = -3
And
2a + (-3) = 47
2a = 50
a = 25

The sum of an arithmetic series to the nth term, S(n), is given by
S(n) = (n/2)[2a + (n-1)d]
S(60) = (60/2)[2(25) + 59(-3)]
     = 30(50 - 177)
     = 30(-127)
     = 3810

2). Alternatively, the sum of an arithmetic series to the nth term can be expressed as
S(n) = (n/2)(a + l)
where l is the last(or nth) term
So if a = -13, l = 99 and S(n) = 1419
1419 = (n/2)(-13 + 99)
1419 = (n/2)(86)
1419 = 43n
n = 1419/43 = 33
also,
l = a + (n-1)d
99 = -13 + 32d
32d = 112
d = 112/32
d = 7/2
There are 33 terms with a common difference of 7/2

I hope i have helped you, i'm always open for further explanation.
Regards.