Question

А) b(6), если b(1) = 125 b(3) = 5 б) b(7), если b(1) = -2/9, b(3) = -2 в) b(1), если b(4) = -1, b(6) = -100 ребят плизззззз засада эти прогрессии

Answer 1

а) b(6), если b(1) = 125 b(3) = 5

b3=b1*q^2      q^2=5/125=1/25

q1=-1/5     b6=b1*q^5=5^3*-5^-5=-1/25

q2=1/5       b6=1/25

б) b(7), если b(1) = -2/9, b(3) = -2

q^2=b3/b1=-2*-9/2=9

q1=-3     q2=3

b7=b1*q^6=-2/9*3^6=-2*3^4=-2*81=-162

в) b(1), если b(4) = -1, b(6) = -100

b6=b4*q^2       q^2=-100/-1=100

q1=-10     b1=b4/q^3=-1/-1000=0.001

q2=10       b1=-0.001

Answer 2

Task A

To find the value of b(6) when b(1) = 125 and b(3) = 5, we can use the formula for the nth term of a geometric progression. The formula is given by:

\[ b(n) = b(1) \times q^{(n-1)} \]

Where: - \( b(n) \) is the nth term - \( b(1) \) is the first term - \( q \) is the common ratio - \( n \) is the term number

Using the given values: - \( b(1) = 125 \) - \( b(3) = 5 \)

We can solve for the common ratio \( q \) using the formula:

\[ b(3) = b(1) \times q^{(3-1)} \] \[ 5 = 125 \times q^2 \] \[ q^2 = \frac{5}{125} \] \[ q^2 = \frac{1}{25} \] \[ q = \pm \frac{1}{5} \]

Now, we can find the value of b(6) using the common ratio \( q = \pm \frac{1}{5} \):

\[ b(6) = 125 \times \left(\pm \frac{1}{5}\right)^{5} \]

The value of \( b(6) \) will depend on the specific value of the common ratio \( q \).

Task B

To find the value of b(7) when b(1) = -2/9 and b(3) = -2, we can use the same formula for the nth term of a geometric progression:

\[ b(n) = b(1) \times q^{(n-1)} \]

Using the given values: - \( b(1) = -\frac{2}{9} \) - \( b(3) = -2 \)

We can solve for the common ratio \( q \) using the formula:

\[ b(3) = b(1) \times q^{(3-1)} \] \[ -2 = -\frac{2}{9} \times q^2 \] \[ q^2 = \frac{2}{9} \] \[ q = \pm \sqrt{\frac{2}{9}} \]

Now, we can find the value of b(7) using the common ratio \( q = \pm \sqrt{\frac{2}{9}} \):

\[ b(7) = -\frac{2}{9} \times \left(\pm \sqrt{\frac{2}{9}}\right)^{6} \]

The value of \( b(7) \) will depend on the specific value of the common ratio \( q \).

Task C

To find the value of b(1) when b(4) = -1 and b(6) = -100, we can use the formula for the nth term of a geometric progression:

\[ b(n) = b(1) \times q^{(n-1)} \]

Using the given values: - \( b(4) = -1 \) - \( b(6) = -100 \)

We can solve for the common ratio \( q \) using the formula:

\[ b(6) = b(1) \times q^{(6-1)} \] \[ -100 = b(1) \times q^5 \]

We can then solve for \( b(1) \) using the value of the common ratio \( q \):

\[ b(1) = -\frac{100}{q^5} \]

The value of \( b(1) \) will depend on the specific value of the common ratio \( q \).