Question

b) If the sum of three consecutive numbers in an arithmetic progression is 39 and their product is
1989, find the three numbers.​

Answer 1

The three terms of an arithmetic progression are (8, 13 and 18) or, (18, 13 and 8).

Step-by-step explanation:

Let three terms of an arithmetic progression = (a - d), a, ( a + d)

To find, the three terms of an arithmetic progression = ?

According to question,

The sum of three terms of an arithmetic progression = 39

∴ (a - d) + a + ( a + d) = 39

⇒ 3a = 39

⇒ a = 13

Also, The product of three terms of an arithmetic progression = 1872

∴ (a - d)a( a + d) = 1872

⇒ a(a^{2}-d)^{2}a

2

−d)

2

= 1872

Put a = 13, we get

⇒ 13(13^{2}-d^2)=187213

2

−d

2

)=1872

⇒ 169 - d^{2}d

2

= 144

⇒ d^{2}d

2

= 169 - 144 = 25

⇒ d = ± 5

∴ a = 13 and d = ± 5

Put a = 13 and d = 5

Three terms of an arithmetic progression = (13 - 5), 13 and ( 13 + 5)

= 8, 13 and 18

Put a = 13 and d = - 5

Three terms of an arithmetic progression = (13 + 5), 13 and ( 13 - 5)

= 18, 13 and 8

Thus, the three terms of an arithmetic progression are (8, 13 and 18) or, (18, 13 and 8).

Answer 2

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