Answer:
[tex]\large\boxed{10.\ \sum\limits_{n=1}^{9}5(2)^{n-1}=2,555}[/tex]
[tex]\large\boxed{11.\ S_7=6,564}[/tex]
[tex]\large\boxed{12. a_1=160}[/tex]
Step-by-step explanation:
10.
The formula of a sum of terms of a geometric sequence:
[tex]S_n=\dfrac{a_1(1-r^n)}{1-r}[/tex]
We have:
[tex]\sum\limits_{n=1}^{9}5(2)^{n-1}[/tex]
Therefore:
[tex]a_n=5(2)^{n-1}\to a_1=5(2)^{1-1}=5(2)^0=5(1)=5[/tex]
[tex]r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=5(2)^{n+1-1}=5(2)^n\\\\r=\dfrac{5(2)^n}{5(2)^{n-1}}=\dfrac{2^n}{2^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=2^{n-(n-1)}=2^{n-n+1}=2^1=2[/tex]
Substitute a₁ = 5, r = 2 and n = 9:
[tex]S_9=\dfrac{5(1-2^9)}{1-2}=\dfrac{5(1-512)}{-1}=-5(-511)=2,555[/tex]
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11.
We have
[tex]a_2=-36,\ a_5=972,\ n=7[/tex]
We know:
[tex]a_n=a_1r^{n-1}[/tex]
Therefore
[tex]a_2=a_1r^{2-1}=a_1r^1=a_1r\\\\a_5=a_1r^{5-1}=a_1r^4\\\\\dfrac{a_5}{a_2}=\dfrac{a_1r^4}{a_1r}=r^{4-1}=r^3[/tex]
Substitute:
[tex]r^3=\dfrac{972}{-36}\\\\r^3=-27\to r=\sqrt[3]{-27}\\\\r=-3[/tex]
Calculate the first term:
[tex]a_2=a_1r\to a_1=\dfrac{a_2}{r}\\\\a_1=\dfrac{-36}{-3}=12[/tex]
Put a₁ = 12, r = -3 and n = 7 to the formula of a sum:
[tex]S_7=\dfrac{12(1-(-3)^7)}{1-(-3)}=\dfrac{12(1-(-2187))}{1+3}=\dfrac{12(1+2187)}{4}=3(2188)\\\\S_7=6564[/tex]
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12.
We have
[tex]n=6,\ S_n=315\to S_6=315,\ a_n=5\to a_6=5,\ r=\dfrac{1}{2}[/tex]
We know:
[tex]a_n=a_1r^{n-1}\to a_6=a_1r^{6-1}=a_1r^5[/tex]
Substitute:
[tex]5=a_1\left(\dfrac{1}{2}\right)^5\\\\5=\dfrac{1}{32}a_1\qquad\text{multiply both sides by 32}\\\\160=a_1\to a_1=160[/tex]
Check for the given sum:
Substitute a₁ = 160, r = 1/2 and n = 6:
[tex]S_6=\dfrac{160\left(1-\left(\frac{1}{2}\right)^6\right)}{1-\frac{1}{2}}=\dfrac{160\left(1-\frac{1}{64}\right)}{\frac{1}{2}}=160\left(\dfrac{64}{64}-\dfrac{1}{64}\right)\left(\dfrac{2}{1}\right)\\\\=160\left(\dfrac{63}{64}\right)(2)=(2.5)(63)(2)=315[/tex]
CORRECT :)
