yoda82
Answered

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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.
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Sagot :

jacob193

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].

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