Question

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

Answer 1

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$.)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

• $a$,
• $a\cdot r$,
• $a \cdot r^2$,
• $a \cdot r^3$,
• $a \cdot r^4$.

First equation:

$a\, r^4 - a = 150$.

Second equation:

$a\, r^3 - a\, r = 48$.

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$.

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$.

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$.

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$.

Simplify and solve for $r$:

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$.

$8\, r^2 - 25\, r + 8 = 0$.

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$.

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$. Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$.

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$.

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$. Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$.

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$.