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Paraphin [41]
3 years ago
5

Show work and explain with formulas.

Mathematics
1 answer:
Mandarinka [93]3 years ago
4 0

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

==========================================

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

==========================================

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 :)

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