Question

The fourth term of a geometric progression is 4 and the seventh term is 32/125. Find the first term, the common ratio and the sum of the first nth terms. Hence, deduce the sum to infinity.​

Answer 1

Answer:

common ratio is 2/5

First term is  125/2

The sum of the first n terms is: $S_n=\frac{125}{2} \, \frac{1-(2/5)^n}{3/5}$

The infinite sum is: $S_\infty =\frac{625}{6}$

Step-by-step explanation:

The 4th term is 4, and the seventh term is 32/125 so we use the definition of nth term of a geometric sequence to create the following two equations:

$a_4=a_1\,*\,r^{4-1} = a_1\,*\,r^3\\4=a_1\,*\,r^3\\and\\a_7=a_1\,*\,r^{7-1} = a_1\,*\,r^6\\\frac{32}{125} = a_1\,*\,r^6$

Now we divide a7 by a4 to get rid of a1 and work on determining the common ratio of the sequence:

$\frac{a_7}{a_4} =\frac{8}{125} =\frac{a_1\,r^6}{a_1\,r^3} =r^3\\then\\r=\sqrt[3]{\frac{8}{125} } =\frac{2}{5}$

So, the common ratio is 2/5

we can now determine the first term:

$a_4=4=a_1\,*\,(2/5)^3\\a_1=\frac{125*4}{8} =\frac{125}{2}$

The sum of the first n terms is given by the formula:

$S_n=\frac{125}{2} \, \frac{1-(2/5)^n}{1-(2/5)} \\S_n=\frac{125}{2} \, \frac{1-(2/5)^n}{3/5}$

and therefore, the infinite sum is:

$S_\infty = a_1\,\frac{1}{1-r} = \frac{125}{2} \,\frac{5}{3} =\frac{625}{6}$