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3 years ago

the first four terms of an arithmetic sequence are 2,a-b, 2a+b+7 and a - 3b, where a and b are constants. Find a and b

Mathematics

1 answer:

jonny [76]3 years ago

7 0

Arithmetic Sequence

The values of a and b is a = 2 and b=3

Step-by-step explanation:

Given the terms of the arithmetic sequence are

2 , a - b , 2a + b + 7, a - 3b

Let the common difference be D

Therefore,

The difference between the first two consecutive terms is

(a – b) – 2 = D ------------------------------( 1 )

The difference between the next two consecutive terms is

D = (2a + b+7) – ( a - b ) ---------------------(2 )

Equating equation 1 and equation 2

⇒ (a – b) -2 =(2a+b+7)-(a-b)

⇒ a – b – 2 = a + 2b +7

⇒ 3b = -9

⇒ b = -3

Similarly

The difference between the next two consecutive terms is

D = (a-3b)-(2a+b+7) ------- (3)

⇒ (a-3b)-(2a+b+7)=(2a+b+7)-(a-b)

⇒ a-3b)-(2a+b+7) -a - 4b -7 === a+2b+7

⇒ 2a = - ( 14 + 6b)

⇒ a = -( 7 + 3b)

⇒ a = - ( 7 – 3*3 )

Thus the value of a = 2

Hence , the values of a and b is a = 2 and b=3

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2 years ago

What is 0.7, 0.755, 5/8 from least to greatest?

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I like to start by turning all my numbers into the same form. You can go either way but I am going to turn the decimals into fractions. so 0.7 is the same as 7/10.

0.755 is like 755/1000 which can be reduced to 151/200 (if you don't understand the convertions let me know i will explaine more.)

Next give all the fractions a common denominator. both 8 and 10 go into 200 so we can use that. 151/200 already has a denominator of 200 so that one is set.
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the last one is 5/8 8 times 25 is 200 so multiply both numbers by 25 to get 125/200

the fractionos we have are 125/200 151/200 and 140/200. Now because they all have the same denominator the one with the lowest numerator is the least number and the one with the highest number is the greatest number.

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140/200 was 0.7
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3 years ago

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Now that you have the price for 1 hour including the initial fee, now you need to find the price for each hour after that. Here's how I did that:

I created a graph that looked like this:

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