Arithmetic Progression problem
- Thread starter Pankaj
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An A.P. has to features:
1) Starting value, usually "a".
2) Common difference, usually "d".
Any term in the series is easily stated: a + (n-1)d
It is seen that n = 1, the first term, is a + (1-1)d = a + (0)d = a+0 = a, as intended.
The sum of n terms takes a little effort.
a + [a+d] + [a+2d] + ... + [a+(n-1)d]
Rearranging
na + [d + 2d + 3d + ... (n-1)d]
More
na + d[1 + 2 + 3 + ... (n-1)]
We have only to add up the values in the square brackets. It is a well-known result that gives:
na + d[(n-1)n/2]
Our trouble is, we have two such progressions with different starting values and differences (potentially different).
Let's just define another one with starting value 'b' and common difference 'e'.
The sum of n such terms is
nb + e[(n-1)n/2]
That's enough of that. All we've done so far is the basic definitions. Now to solve the problem.
We need a ratio. Show that and see where it leads. Let's see what you get.
1) Starting value, usually "a".
2) Common difference, usually "d".
Any term in the series is easily stated: a + (n-1)d
It is seen that n = 1, the first term, is a + (1-1)d = a + (0)d = a+0 = a, as intended.
The sum of n terms takes a little effort.
a + [a+d] + [a+2d] + ... + [a+(n-1)d]
Rearranging
na + [d + 2d + 3d + ... (n-1)d]
More
na + d[1 + 2 + 3 + ... (n-1)]
We have only to add up the values in the square brackets. It is a well-known result that gives:
na + d[(n-1)n/2]
Our trouble is, we have two such progressions with different starting values and differences (potentially different).
Let's just define another one with starting value 'b' and common difference 'e'.
The sum of n such terms is
nb + e[(n-1)n/2]
That's enough of that. All we've done so far is the basic definitions. Now to solve the problem.
We need a ratio. Show that and see where it leads. Let's see what you get.