The sum of the first three terms of a geometric sequence of positive integers is equal to seven times the first term and the sum of the first four terms is forty five. What is the first term of the sequence.
Here is the solution to this question:
Let us assume that the first term is 'a' and 'r' is the common ratio in this geometric sequence.
With this, first four terms of this sequence will be
a, ar, ar^2, ar^3
Now, Sum of the first three terms is equal to seven times the first term. This means that
a + ar + ar^2 = 7a ------------------ (1)
Dividing both sides by 'a',
1 + r + r^2 = 7
r^2 + r - 6 = 0
Solving this quadratic equation for r, we get,
r+3 = 0 or r-2=0
r = -3 or r=2
However, since this is a sequence of positive integers, r can not be negative. Hence we rule out possibiility of r=-3.
Thus r = 2 ------------------- (2)
Now, sum of first four terms is 45. We write,
a + ar + ar^2 + ar^3 = 45 -----------(3)
However, from equation (1) above, we substitute 'a + ar + ar^2' with 7a in equation (3). This becomes,
7a + ar^3 = 45
From (2) above, putting r = 2,
7a + a*2^3 = 45
7a + 8a = 45
15a = 45
a = 3
a is the first term of our sequence. Hence we get first term of the sequence as 3.
I hope that this helps.
Please let me know if you need additional information. Here I have assumed that you know how to find roots of the quadratic equation. If you want to know details of the same, then please let me know.
All the best!