An arithmetic progression has first term a and common difference -1.?

Question

The sum of the first n terms is equal to the sum of the first 3n terms. Express a in terms of n.

Who can help me with this maths problem?

Answer 1

This is a very interesting question indeed.
An arithmetic progression has first term a and common difference -1.

If the initial term of an arithmetic progression is a1 and the common difference of successive members is d, then the nth term of the sequence is given by:
an = a1 + (n - 1)d;
d = -1 and a1 = a, so an = a - (n - 1);

The value of an arithmetic series consisting of n terms a1, a2, a3,...,an is given by
Sn = a1 + a2 + ... + an = n(a1 + an)/2 = n[2a1 + (n - 1)d]/2
in this case, Sn = n[2a - (n - 1)]/2

If the sum of the first n terms is equal to the sum of the first 3n terms, then Sn = S3n.
Sn = n[2a - (n - 1)]/2;
S3n = 3n[2a - (3n - 1)]/2;
n[2a - (n - 1)]/2 = 3n[2a - (3n - 1)]/2;
Simplify,
2a - n + 1 = 3(2a - 3n + 1);
2a - n + 1 = 6a - 9n + 3;
4a = 8n - 2;
a = 2n - 1/2;

Answer 2

a(k) = a - k, k is an integer >= 0

Each sequence can be viewed as a postively increasing sequence with the last term as the first and the first term as the last. So then you can use the summation formula S = (first + last)/2 x (number of terms).

The first n terms would be a - (n - 1) to a:

n(a - (n - 1) + a)/2 = n/2(-n + 1 + 2a)

The first 3n terms would be a - (3n - 1) to a:

3n(a - (3n - 1) + a)/2 = 3n/2(-3n + 1 + 2a)

Set them equal:

n/2(-n + 1 + 2a) = 3n/2(-3n + 1 + 2a)
-n + 1 + 2a = -9n + 3 + 6a
8n - 2 = 4a
a = (8n - 2)/4 = 2n - 1/2

Answer 3

a, a-1, a-2, a-3...

nth term = a + (n-1)d
= a - n + 1

sum of first n terms = n/2 (a + l)
=n/2 (a + a - n + 1)

sum of first 3n terms = 3n/2 (a + a - 3n + 1)

sum of first n terms = sum of first 3n terms
n/2 (2a - n + 1) = 3n/2(2a - 3n + 1)
2a - n + 1 = 6a - 9n +3
4a = 8n - 2
a = 2n - 1/2

Answer 4

Sn=n/2[2a+(n-1)(-1)]
Sn=n/2[2a-n+1]

S3n=3n/2[2a+(3n-1)(-1)]
S3n=3n/2[2a-3n+1]

Sn=S3n
n/2[2a-n+1]=3n/2[2a-3n+1]
2a-n+1=6a-9n+3
8n=4a+2
a=2n-1/2