Question

Show work and explain with formulas.

Answer 1

Answer:

$\large\boxed{10.\ \sum\limits_{n=1}^{9}5(2)^{n-1}=2,555}$

$\large\boxed{11.\ S_7=6,564}$

$\large\boxed{12. a_1=160}$

Step-by-step explanation:

10.

The formula of a sum of terms of a geometric sequence:

$S_n=\dfrac{a_1(1-r^n)}{1-r}$

We have:

$\sum\limits_{n=1}^{9}5(2)^{n-1}$

Therefore:

$a_n=5(2)^{n-1}\to a_1=5(2)^{1-1}=5(2)^0=5(1)=5$

$r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=5(2)^{n+1-1}=5(2)^n\\\\r=\dfrac{5(2)^n}{5(2)^{n-1}}=\dfrac{2^n}{2^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=2^{n-(n-1)}=2^{n-n+1}=2^1=2$

Substitute a₁ = 5, r = 2 and n = 9:

$S_9=\dfrac{5(1-2^9)}{1-2}=\dfrac{5(1-512)}{-1}=-5(-511)=2,555$

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11.

We have

$a_2=-36,\ a_5=972,\ n=7$

We know:

$a_n=a_1r^{n-1}$

Therefore

$a_2=a_1r^{2-1}=a_1r^1=a_1r\\\\a_5=a_1r^{5-1}=a_1r^4\\\\\dfrac{a_5}{a_2}=\dfrac{a_1r^4}{a_1r}=r^{4-1}=r^3$

Substitute:

$r^3=\dfrac{972}{-36}\\\\r^3=-27\to r=\sqrt[3]{-27}\\\\r=-3$

Calculate the first term:

$a_2=a_1r\to a_1=\dfrac{a_2}{r}\\\\a_1=\dfrac{-36}{-3}=12$

Put a₁ = 12, r = -3 and n = 7 to the formula of a sum:

$S_7=\dfrac{12(1-(-3)^7)}{1-(-3)}=\dfrac{12(1-(-2187))}{1+3}=\dfrac{12(1+2187)}{4}=3(2188)\\\\S_7=6564$

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12.

We have

$n=6,\ S_n=315\to S_6=315,\ a_n=5\to a_6=5,\ r=\dfrac{1}{2}$

We know:

$a_n=a_1r^{n-1}\to a_6=a_1r^{6-1}=a_1r^5$

Substitute:

$5=a_1\left(\dfrac{1}{2}\right)^5\\\\5=\dfrac{1}{32}a_1\qquad\text{multiply both sides by 32}\\\\160=a_1\to a_1=160$

Check for the given sum:

Substitute a₁ = 160, r = 1/2 and n = 6:

$S_6=\dfrac{160\left(1-\left(\frac{1}{2}\right)^6\right)}{1-\frac{1}{2}}=\dfrac{160\left(1-\frac{1}{64}\right)}{\frac{1}{2}}=160\left(\dfrac{64}{64}-\dfrac{1}{64}\right)\left(\dfrac{2}{1}\right)\\\\=160\left(\dfrac{63}{64}\right)(2)=(2.5)(63)(2)=315$

CORRECT :)