Question

(a) Find the value of xif 10, x and 30 are in A.P.

Answer 1

Part A

Answer:  x = 20

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Explanation:

AP = arithmetic progression, which is another way of saying arithmetic sequence.

Let d = common difference

• first term = 10
• second term = first+d = 10+d = x
• third term = second+d = x+d = 30

Apply substitution like so

x+d = 30

10+d+d = 30 ... replaced x with 10+d

10+2d = 30

2d = 30-10

2d = 20

d = 20/2

d = 10 is the common difference

Therefore,

x = 10+d = 10+10 = 20

The AP {10,x,30} updates to {10,20,30}. We see the gap between terms is 10 units. All AP's have the same gap width between adjacent terms.

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Part B

Answers:  x = 10 and y = 14

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Explanation:

We use the same ideas mentioned back in part A.

d = common difference = unknown for now

• first term = 6
• second term = first+d = 6+d = x
• third term = second+d = x+d = (6+d)+d = 6+2d = y
• fourth term = third+d = y+d = (6+2d)+d = 6+3d = 18

Hopefully you can see how each term builds up to form the next one. We'll solve that last equation like so

6+3d = 18

3d = 18-6

3d = 12

d = 12/3

d = 4

So we then can say:

• x = 6+d = 6+4 = 10
• y = 6+2d = 6+2(4) = 6+8 = 14

The arithmetic sequence {6,x,y,18} updates to {6,10,14,18}. There's a gap of 4 between each adjacent term.