Question

The sum of first and second term of a geometric progression is 3 and the sum of

3rd and 4th terms is 12. Find the first term and the common ratio of the progression
and state the first four terms of the progression.

Answer 1

Let the first and second term be a and r respectively.

The terms up to the 4th term will be: a, ar, ar^2, ar^3.

The sum of the first and second terms will, therefore, be a+ar=3 simplified as a(1+r)=3

The sum of the third and the fourth term will, therefore, be ar^2 + ar^3=12 simplified as ar^2 (1+r)=12

As seen, the common factor is (1+r) and this easily calls for the changing of both equations to make this the common factor. The equations will, therefore, be (1+r)=3a and (1+r)=12ar^2

We proceed to equate these equations as a simplification to find the unknowns. This will be 3a=12ar^2 that breaks down to 1/4 = r^2

Finding the square root of both sides gives the answer that the common ration r =1/2 and also r=-1/2.

If r=1/2 then,

According to the formula a+ar=3, changes to after replacing r to get a+1/2a=3, a= 2

The first four terms will be a, ar, ar^2, ar^3, which is 2, (2*1/2), (2*1/2^2), (2*1/2^3), which is 2, 1, 1/2, 1/4

The Answers are, therefore:

The First Term is 2

The Common Ratio is 1/2

The First Four Terms are 2, 1, 1/2, 1/4

On the other hand, if r =-1/2 then using the equation a(1+r)=3, then a=3/(1+-1/2), which is a=3/0.5=6

If r = -1/2 then a=6

The first four terms will therefore be, a, ar, ar^2, ar^3, which is 6, -3, 1.5 and -3/4.

If r=-1/2 the Answers are, therefore:

The First Term is 6

The Common Ratio is is r as -1/2

The First Four Terms are 6, -3, 1.5, -3/4