+0

# geometric progression

0
93
3

In a geometric progression of real numbers, the sum of the first two terms is 7, and the sum of the first six term is 91.  Find the sum of the first four terms.

Feb 18, 2020

#1
+6180
+2

$$a \dfrac{1-r^6}{1-r} = 91 = 13 \cdot 7 = 13 a\dfrac{1-r^2}{1-r}\\ \dfrac{1-r^6}{1-r^2}=13\\ r^6-13r^2+12=0\\ r=\pm 1,~\pm \sqrt{3}\\ \text{We only need consider r=\sqrt{3}}\\ a \dfrac{1-3}{1-\sqrt{3}}=7\\ a = \dfrac 7 2 (\sqrt{3}-1)\\ a \dfrac{1-r^4}{1-r} = 28$$

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Feb 18, 2020
#2
+109740
+1

(1 - r^2)                                          (1 - r^6)

a*_______  =  7          and      a  *  _________  =  91

1 - r                                            (1   - r)

a * (1  -r^2) =  7 (1 - r)                a ( 1 -r^6)  =  91 ( 1 - r)

13 a ( 1 - r^2) = 91 (1 - r)          a ( 1 - r^6)  = 91 ( 1 - r)

So

13a ( 1 -r^2)  =  a ( 1 - r^6)

13 ( 1 - r^2)  = ( 1 -r^6)

13 - 13r^2 = 1 - r^6

r^6  - 13r^2  + 12  =   0

Note that   r = 1   and  r  = -1  are roots

So  (r + 1) ( r -1)  =  r^2  - 1

Dividing we have

r^4  + r^2  - 12

r^2  -1   [  r^6  - 13r^2 +  12]

r^6                          - 1r^4

_____________________________________

r^4  -13r^2 + 12

r^4 -  r^2

_______________

-12r^2  + 12

-12r^2  +  12

____________

0

The remaining polynomial  =   r^4  + r^2  - 12

Set to 0  and factor

(r^2 + 4) ( r^2 - 3)   = 0

Since the  sum of the  first n terms increases, then  √3  is the only possible value  for  r

So

7 =  a ( 1 - 3) / ( 1  -√3)

7 = -2a / (1 - √3)

7=  2a / (√3 - 1)

a = (7/2)(√3 - 1)

And the sum of the 1st  4 terms   =

(7/2) (√3 -1)  ( 1 - 9) / ( 1 - √3)  =

(7/2) (√3  -1) (-8) / ( 1 - √3)   =

(7/2) (√3 -1) (8) / ( √3 - 1)  =

(7/2) (8)  =

28

Feb 18, 2020
#3
+24430
+1

In a geometric progression of real numbers, the sum of the first two terms is 7,

and the sum of the first six term is 91.

Find the sum of the first four terms.

$$\begin{array}{|rclrcl|} \hline \mathbf{s_4 = a \dfrac{1-r^4}{1-r}} && \mathbf{s_2 = a \dfrac{1-r^2}{1-r}} \\ \hline \\ \dfrac{s_4}{s_2} &=& a \dfrac{1-r^4}{1-r} \above 1pt a \dfrac{1-r^2}{1-r} \\\\ \dfrac{s_4}{s_2} &=& \dfrac{(1-r^4)}{1-r^2} \\\\ \dfrac{s_4}{s_2} &=& \dfrac{(1-r^2)(1+r^2)}{1-r^2} \\\\ \dfrac{s_4}{s_2} &=& 1+r^2 \\\\ s_4 &=& s_2(1+r^2) \quad | \quad s_2 = 7 \\\\ \mathbf{s_4} &=& \mathbf{7(1+r^2)} \qquad (1) \\ \hline \end{array} \begin{array}{|rclrcl|} \hline \mathbf{s_2} &=& \mathbf{a+ ar} \\\\ s_2 &=& a(1+r) \quad | \quad s_2 = 7 \\\\ 7 &=& a(1+r) \\\\ \mathbf{a} &=& \mathbf{\dfrac{7}{1+r} } \qquad (2) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline s_6 &=& S_2 + ar^2+ar^3+ar^4+ar^5 \quad | \quad s_2 = 7, s_6 = 91 \\ 91 &=& 7 + ar^2+ar^3+ar^4+ar^5 \\ 84 &=& ar^2+ar^3+ar^4+ar^5 \\ 84 &=& ar^2 (1 +r+r^2+r^3) \quad | \quad \mathbf{a=\dfrac{7}{1+r} } \\\\ 84 &=& \dfrac{7r^2(1 +r+r^2+r^3)}{1+r} \quad | \quad 1 +r+r^2+r^3 = \dfrac{1-r^4}{1-r} \\\\ 84 &=& \dfrac{7r^2\dfrac{1-r^4}{1-r}}{1+r} \\\\ 84 &=& \dfrac{7r^2(1-r^4)}{(1-r)(1+r)} \\\\ 84 &=& \dfrac{7r^2(1-r^4)}{(1-r^2)} \\\\ 84 &=& \dfrac{7r^2(1-r^2)(1+r^2)}{(1-r^2)} \\\\ 84 &=& 7r^2(1+r^2) \quad | \quad : 7 \\\\ 12 &=& r^2(1+r^2) \\\\ 12 &=& r^2 + r^4 \\\\ r^4+r^2 -12 &=& 0 \\\\ r^2 &=& \dfrac{-1\pm \sqrt{1-4*(-12)} }{2} \\ r^2 &=& \dfrac{-1\pm \sqrt{49} }{2} \\ r^2 &=& \dfrac{-1\pm 7 }{2} \\ \\ r^2 &=& \dfrac{-1+7 }{2} \\ r^2 &=& \dfrac{6 }{2} \\ r^2 &=& \mathbf{3} \\\\ r^2 &=& \dfrac{-1- 7 }{2} \\ r^2 &=& \dfrac{-8 }{2} \\ r^2 &=& -4 \qquad \text{r is a complex number! No solution} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{s_4} &=& \mathbf{7(1+r^2)} \quad | \quad r^2 = 3 \\ s_4 &=& 7(1+3) \\ s_4 &=& 7*4 \\ \mathbf{s_4} &=& \mathbf{28} \\ \hline \end{array}$$

The sum of the first four terms is 28

Feb 19, 2020