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

2nd and the 4th terms is 48. Find the sum of the first five terms.


Sagot :

Answer:

Either [tex]\displaystyle \frac{-1522}{\sqrt{41}}[/tex] (approximately [tex]-238[/tex]) or [tex]\displaystyle \frac{1522}{\sqrt{41}}[/tex] (approximately [tex]238[/tex].)

Step-by-step explanation:

Let [tex]a[/tex] denote the first term of this geometric series, and let [tex]r[/tex] denote the common ratio of this geometric series.

The first five terms of this series would be:

  • [tex]a[/tex],
  • [tex]a\cdot r[/tex],
  • [tex]a \cdot r^2[/tex],
  • [tex]a \cdot r^3[/tex],
  • [tex]a \cdot r^4[/tex].

First equation:

[tex]a\, r^4 - a = 150[/tex].

Second equation:

[tex]a\, r^3 - a\, r = 48[/tex].

Rewrite and simplify the first equation.

[tex]\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}[/tex].

Therefore, the first equation becomes:

[tex]a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150[/tex]..

Similarly, rewrite and simplify the second equation:

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

Therefore, the second equation becomes:

[tex]a\, r\, \left(r^2 - 1\right) = 48[/tex].

Take the quotient between these two equations:

[tex]\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}[/tex].

Simplify and solve for [tex]r[/tex]:

[tex]\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}[/tex].

[tex]8\, r^2 - 25\, r + 8 = 0[/tex].

Either [tex]\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}[/tex] or [tex]\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}[/tex].

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

Calculate the sum of the first five terms:

[tex]\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}[/tex].

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

Calculate the sum of the first five terms:

[tex]\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}[/tex].