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3 years ago

The 4th and 8th term of a G.P. are 24 and 8/27 respectively. find the 1st term and common ratio

Mathematics

1 answer:

beks73 [17]3 years ago

4 0

Answer:

see explanation

Step-by-step explanation:

The n th term of a geometric progression is

• $a_{n}$ = a₁ $r^{n-1}$

where a₁ is the first term and r the common ratio

given a₄ = 24, then

a₁ $r^{3}$ = 24 → (1)

Given a₈ = $\frac{8}{27}$ , then

a₁ $r^{7}$ = $\frac{8}{27}$ → (2)

Divide (2) by (1)

$r^{4}$ = $\frac{\frac{8}{27} }{24}$ = $\frac{1}{81}$

Hence r = $\sqrt[4]{\frac{1}{81} }$ = $\frac{1}{3}$

Substitute this value into (1)

a₁ × ( $\frac{1}{3}$ )³ = 24

a₁ × $\frac{1}{27}$ = 24, hence

a₁ = 24 × 27 = 648

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2 years ago

An electronics company produces transistors, resistors, and computer chips. Each transistor requires 3 units of copper, 1 unit

padilas [110]

Answer:

An electronics company can be produce 350 transistors and 340 computer chips, they can´t produce resistors.

Step-by-step explanation:

1. We will name the variables for transistors, resistors and the computer chips.

a = Transistors

b= Resistors

c = Computer chips

2. We propose three linear equations, one for the copper, one for the zinc and one for the glass.

$\left \{ {{3a+3b+2c=1730} \atop {a+2b+c=690}}\atop {2a+b+2c=1380}} \right.$

3. We write the matrix form as Ax=d

$A=\left(\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right)$

$x=\left(\begin{array}{ccc}a\\b\\c\end{array}\right)$

$A=\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)$

With this formula the solution of x is $x=\frac{d}{A}$ or $x=A^{-1}d$

4. We will find the inverse matrix $A^{-1}$ using the formula:

$A^{-1} = \frac{1}{detA} (C_{A})^{T}$

a. det A

$det A=\left[\begin{array}{ccc}3&3&2\\1&2&1\\2&1&2\end{array}\right] =3*(4-1)-3*(2-2)+2*(1-4)=9-0-6=3$

b. $(C_{A})^{T}$

$C_{A}=\left(\begin{array}{ccc}4-1&.(2-2)&1-4\\-(6-2)&6-4&-(3-6)\\3-4&-(3-2)&6-3\end{array}\right)$

$C_{A}=\left(\begin{array}{ccc}3&.0&-3\\-4&2&3\\-1&-1&3\end{array}\right)$

$(C_{A}) ^T=\left(\begin{array}{ccc}3&0&-3\\-4&2&3\\-1&-1&3\end{array}\right)^T$

$(C_{A}) ^T=\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)$

c. $A^{-1}$

$A^{-1}=\frac{1}{3} \left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)$

5. As $x=\frac{d}{A}$ or $x=A^{-1}d$ , the solution of x is:

$x=\frac{1}{3}\left(\begin{array}{ccc}3&-4&-1\\0&2&-1\\-3&3&3\end{array}\right)\left(\begin{array}{ccc}1730\\690\\1380\end{array}\right)$

$x=\frac{1}{3}\left(\begin{array}{ccc}(3*1730)+(-4*690)+(-1*1380)\\(0*1730)+(2*690)+(-1*1380)\\(-3*1730)+(3*690)+(3*1380)\end{array}\right)$

$x=\frac{1}{3}\left(\begin{array}{ccc}1050\\0\\1020)\end{array}\right)$

$X=\left[\begin{array}{ccc}350\\0\\340\end{array}\right]$

Therefore:

a= 350 Transistors

b=0 Resistors

c=340 Computer chips

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2 years ago