Arithmetics Question C1?

Question

Pls help me out here! I have a maths exam 2moro and I need you to help me with this question. I'll copy it exactly:

Three consecutive terms of an arithmetic series have sum of 21 and product of 315. Find the numbers (a-d), a, (a+d)
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Note: a => first term
d => common difference

Answer 1

Since the sum of the terms is 21, we have

(a-d) + a + (a+d) = 3a = 21,

and so a = 7.

Similarly, since their produce is 315, we have

(a-d)*a*(a+d) = 7*(49 - d^2) = 315,

and so 49 - d^2 = 45, ie. d = 2 or -2.

Hence, the consecutive terms are 5, 7 and 9.

Answer 2

I assume you mean even or odd terms.

let

x = first term

x + 2 = second term

x + 4 = Third term

21 = the sum of the terms

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x + x + 2 + x + 4 = 21

3x + 6 = 21

3x + 6 - 6 = 21 - 6

3x = 15

3x / 3 = 15 / 3

x = 15 / 3

x = 5. . . . .The first term

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x = 5

x + 2 = 5 + 2 = 7 The second term

x + 4 = 5 + 4 = 9 The third term

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The product

5 x 7 x 9 = 315

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Answer 3

a=7
d=2

(a-d)+a+(a+d)=3a=21

(a-d)(a+d)a=(a^2-d^2)a=
a^3-ad^2=343-7d^2=315
7d^2=28