+116 The sum of a geometric series whose first three terms are 8000, -12000, and 18000 is 57875. What is the last term of the series?
A geometric series \(b_1+b_2+b_3+\cdots+b_{10}\) has a sum of 180. Assuming that the common ratio of that series is \(\dfrac{7}{4}\), find the sum of the series \(b_2+b_4+b_6+b_8+b_{10}.\)
+80 2) b + bx +bx^2 +bx^3 .... +bx^9 = 180, where x =7/4
Therefore, b= 47185920/93808891
Therefore, b_2+b_4+b_6+b_8+b_10 = 1260/11.
1) By the geometric series (finite) formula,
\(\frac{8000(1-(-1.5)^n)}{2.5}=57875 \\ \text{where n is the number of terms}\)
n=7, so 8000 times -1.5^7 = -136687.5.
+26367 The sum of a geometric series whose first three terms are 8000, -12000, and 18000 is 57875.
What is the last term of the series?
Geometric series:
\(\begin{array}{lrcll} \text{first term} & a_1 &=& a \\ \text{second term} & a_2 &=& ar \\ \text{third term} & a_3 &=& ar^2 \\ \ldots \\ \text{last term} & a_n &=& ar^{n-1} \\\\ \text{so}& a_1 &=& 8000 \\ & a_2 &=& -12000 \\ & a_3 &=& 18000 \\\\ \text{a=?} & a &=& a_1 \\ & \mathbf{a} & \mathbf{=}& \mathbf{8000} \\\\ \text{r=?} & r &=& \dfrac{a_2}{a_1} = \dfrac{a_3}{a_2} \\ & r &=& \dfrac{-12000}{8000} \\ & \mathbf{r} & \mathbf{=}& \mathbf{-\dfrac{3}{2}} \\ \end{array} \)
The sum of a geometric series \(s_n\):
\(\begin{array}{|rcll|} \hline s_n &=& a\left(\dfrac{1-r^{n}}{1-r}\right) \\\\ s_n &=& \dfrac{a-ar^{n}}{1-r} \\\\ s_n &=& \dfrac{a-ar^{n-1}r}{1-r} \quad | \quad a_n = ar^{n-1} \\\\ \mathbf{s_n} & \mathbf{=}& \mathbf{\dfrac{a-a_nr}{1-r} } \\ \hline \end{array}\)
The last term \(a_n\):
\(\begin{array}{|rcll|} \hline s_n &=& \dfrac{a-a_nr}{1-r} \\ s_n(1-r) &=& a-a_nr \\ a_nr &=& a-s_n(1-r) \\ a_n &=& \dfrac{ a-s_n(1-r) } {r} \quad & | \quad a=8000,\ s_n =57875, \ r =-\dfrac{3}{2} \\ a_n &=& \dfrac{ 8000-57875\left(1-\left(-\dfrac{3}{2}\right)\right) } {-\dfrac{3}{2}} \\ a_n &=& -\dfrac{2}{3}\cdot \left( 8000-57875\cdot \dfrac{5}{2} \right) \\ a_n &=& \dfrac{1}{3}\cdot \left( 5\cdot 57875-16000 \right) \\ a_n &=& \dfrac{273375}{3} \\ \mathbf{a_n} & \mathbf{=}& \mathbf{91125} \\ \hline \end{array}\)
The last term of the series is 91125
+26367 A geometric series
\(b_1+b_2+b_3+\cdots+b_{10} \) has a sum of 180.
Assuming that the common ratio of that series is \(\dfrac{7}{4}\),
find the sum of the seres \(b_2+b_4+b_6+b_8+b_{10}\).
Geometric series:
\(\begin{array}{|rcll|} \hline b_1 &=& a \\ b_2 &=& ar \\ b_3 &=& ar^2 \\ b_4 &=& ar^3 \\ b_5 &=& ar^4 \\ b_6 &=& ar^5 \\ b_7 &=& ar^6 \\ b_8 &=& ar^7 \\ b_9 &=& ar^8 \\ b_{10} &=& ar^9 \\ \mathbf{r} &=& \mathbf{ \dfrac{7}{4}} \\\\ s_{10} & = & b_1+b_2+b_3+b_4+b_5+b_6+b_7+b_8+b_9+b_{10} \\ \mathbf{s_{10}} &\mathbf{=}& \mathbf{180} \\ \hline \end{array}\)
The sum of a geometric series \(s_{10}\):
\(\begin{array}{|rcll|} \hline s_{10} &=& a\left(\dfrac{1-r^{10}}{1-r}\right) \\ &\text{or} \\ \mathbf{a} & \mathbf{=}& \mathbf{ \dfrac{s_{10}(1-r) }{1-r^{10}} } \\ \hline \end{array} \)
\(\text{Let $\mathbf{x}=b_2+b_4+b_6+b_8+b_{10}$}\)
\(\begin{array}{|rcll|} \hline s_{10} &=& b_1+b_2+b_3+b_4+b_5+b_6+b_7+b_8+b_9+b_{10} \\\\ s_{10} &=& b_1+(b_2+b_4+b_6+b_8+b_{10})+b_3+b_5+b_7+b_9 \\\\ s_{10} &=& b_1+(b_2+b_4+b_6+b_8+b_{10})+r\left(\dfrac{b_3}{r}+\dfrac{b_5}{r}+\dfrac{b_7}{r}+\dfrac{b_9}{r}\right) \\\\ s_{10} &=& b_1+(b_2+b_4+b_6+b_8+b_{10})+r\cdot (b_2+b_4+b_6+b_8) \\\\ s_{10} &=& b_1+(b_2+b_4+b_6+b_8+b_{10})+r\cdot (b_2+b_4+b_6+b_8)+rb_{10}-rb_{10} \\\\ s_{10} &=& b_1+(b_2+b_4+b_6+b_8+b_{10})+r\cdot (b_2+b_4+b_6+b_8+b_{10})-rb_{10} \\\\ s_{10} &=& b_1+x+r\cdot x-rb_{10} \\\\ s_{10} &=& x(1+r) + b_1-rb_{10} \quad | \quad b_1 = a,\ b_{10}= ar^9 \\\\ s_{10} &=& x(1+r) + a-rar^9 \\\\ s_{10} &=& x(1+r) + a-ar^{10} \\\\ s_{10}&=& x(1+r) + a(1-r^{10}) \quad | \quad \mathbf{a = \dfrac{s_{10}(1-r) }{(1-r^{10})} } \\\\ s_{10}&=& x(1+r) + \dfrac{s_{10}(1-r) }{(1-r^{10})}(1-r^{10}) \\\\ \mathbf{s_{10}} &\mathbf{=}& \mathbf{x(1+r) + s_{10}(1-r) } \\ \hline \end{array} \)
\(\mathbf{x=\ ?}\)
\(\begin{array}{|rcll|} \hline \mathbf{s_{10}} &\mathbf{=}& \mathbf{x(1+r) + s_{10}(1-r) } \\\\ x(1+r) &=& s_{10} - s_{10}(1-r) \\ x(1+r) &=& s_{10}\Big(1 - (1-r) \Big) \\ x(1+r) &=& s_{10} (1 - 1+r ) \\ x(1+r) &=& s_{10}r \\ \mathbf{x} &\mathbf{=}& \mathbf{s_{10}\left(\dfrac{r}{1+r}\right)} \quad | \quad s_{10}=180,\ r=\dfrac{7}{4} \\ x & = & 180\cdot \left(\dfrac{\dfrac{7}{4}}{1+\dfrac{7}{4}}\right) \\\\ x & = & 180\cdot \left(\dfrac{\dfrac{7}{4}}{ \dfrac{11}{4}}\right) \\\\ x & = & 180\cdot \left( \dfrac{7}{11} \right) \\\\ \mathbf{x} &\mathbf{=}& \mathbf{\dfrac{1260}{11}} \\ \hline \end{array}\)
The sum of the series \(\mathbf{b_2+b_4+b_6+b_8+b_{10}}\) is \(\mathbf{\dfrac{1260}{11}}\)
