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fredd [130]
2 years ago
5

In ∆ KLM above, NO // ML and < KNO = < KON. Find m < MNL =

Mathematics
1 answer:
4vir4ik [10]2 years ago
6 0

Answer: In ∆ KLM above, NO // ML and < KNO = < KON. Find m < MNL =

Step-by-step explanation: dUNNO

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Someeee one?????????????????
Ad libitum [116K]

Answer:

Option B) a_{n} = 2\cdot 4^{n-1}

Step-by-step explanation:

The given geometric sequence is

2, 8, 32, 128,....

The general form of a geometric sequence is given by

a_{n} = a_{1}\cdot r^{n-1}

Where n is the nth term that we want to find out.

a₁ is the first term in the geometric sequence that is 2

r is the common ratio and can found by simply dividing any two consecutive numbers in the sequence,

r=\frac{8}{2} = 4

You can try other consecutive numbers too, you will get the same common ratio

r=\frac{32}{8} = 4

r=\frac{128}{32} = 4

So the common ratio is 4 in this case.

Substitute the value of a₁ and r into the above general equation

a_{n} = 2\cdot 4^{n-1}

This is the general form of the given geometric sequence.

Therefore, the correct option is B

Note: Don't multiply the first term and common ratio otherwise you wont get correct results.

Verification:

a_{n} = 2\cdot 4^{n-1}

Lets find out the 2nd term

Substitute n = 2

a_{2} = 2\cdot 4^{2-1} = 2\cdot 4^{1} = 2\cdot 4 = 8

Lets find out the 3rd term

Substitute n = 3

a_{3} = 2\cdot 4^{3-1} = 2\cdot 4^{2} = 2\cdot 16 = 32

Lets find out the 4th term

Substitute n = 4

a_{4} = 2\cdot 4^{4-1} = 2\cdot 4^{3} = 2\cdot 64 = 128

Lets find out the 5th term

Substitute n = 5

a_{5} = 2\cdot 4^{5-1} = 2\cdot 4^{4} = 2\cdot 256 = 512

Hence, we are getting correct results!

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