Part (a): Find the sum
a+(a+1)+(a+2)+.....+(a+n-1)
in terms of a and n
Part (b): Find all pairs of positive integers (a,n) such that n>= 2 and
a+a+1)+(a+2)+....+(a+n-1)=100.
a+(a+1)+(a+2)+.....+(a+n-1)
looks like an arithmetic series with
first term a
common difference = a+1 - a = 1
for n terms
so sum = (n/2) (2a + (n-1)(1))
= (n/2)(2a + n-1)
so (n/2)(2a + n - 1) = 100
2an + n^2 - n = 200
2an = 200 - n^2 + n
a = (200 - n^2 + n)/(2n)
n .. a
2 - 49.5
3 - 32.33..
4 - 23.5
5 - 18
6 - 14.1666..
7 - 11.28...
8 - 9 ---------- yeah
9 - not integer
10 - not integer
11 - not integer
12 - not integer
13 - not integer
14 - .642..
15 - negative
testing for (9,8)
a = 9 , d = 1, and n = 8
sum(8) = (8/2)(18 + 7) = 100
Part (a):
To find the sum of the sequence a+(a+1)+(a+2)+.....+(a+n-1) in terms of a and n, we can apply the formula for the sum of an arithmetic series.
The sum of an arithmetic series is given by 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 our case, the first term is a, the number of terms is n, and the common difference is 1 (as we are adding consecutive terms).
Applying the formula, we have:
S = (n/2)(2a + (n - 1)(1))
S = (n/2)(2a + n - 1)
S = n(a + (n - 1)/2)
So, the sum of the sequence a+(a+1)+(a+2)+.....+(a+n-1) in terms of a and n is S = n(a + (n - 1)/2).
Part (b):
To find all pairs of positive integers (a, n) such that n >= 2 and the sum of the arithmetic series (a+(a+1)+(a+2)+.....+(a+n-1)) is equal to 100, we can set up an equation and solve it.
Using the formula from part (a), we have:
100 = n(a + (n - 1)/2)
Simplifying the equation, we get:
200 = n(2a + n - 1)
We now need to find all pairs (a, n) that satisfy this equation.
One approach is to test different values of n (starting from 2) and solve for a. Since a and n should be positive integers, we can try integer values for n and calculate the corresponding value of a.
Here is a step-by-step process to find all pairs (a, n) that satisfy the equation:
1. Start with n = 2.
2. Substitute n = 2 into the equation: 200 = 2(2a + 2 - 1)
Simplify and solve for a: 200 = 2(2a + 1) => 200 = 4a + 2 => 4a = 198 => a = 49.5 (not a positive integer)
3. Increment n by 1.
4. Repeat step 2 and 3 until you find positive integer solutions for a.
By following this process, you can find all possible pairs (a, n) that satisfy the equation.