Question

The sum of three numbers in an AP in 54 and the product of the first and the third is 288. Find the numbers.​

Answer 1

Answer:

Let the 3 consecutive numbers in AP be (n-d), n and (n+d) where ‘d’ is the common difference.

Since the sum of 3 consecutive numbers is 27,

(n-d) + n + (n+d) = 27

3n = 27

So, n = 9.

The product of 3 consecutive numbers is 504, so,

(n-d) × n × (n+d) = 504

(n^2-d^2) × n = 504

Substitute n = 9 here,

(81-d^2)×9 = 504

81 - d^2 = 504/9

81 -d^2 = 56

-d^2 = 56–81

-d^2 = -25

d^2 = 25

d = 5 or d = -5

Now substitute 'n' and 'd' values in the arithmetic sequence we assumed.

The possible Arithmetic Progressions are:

4, 9, 14 and

14, 9, 4.

Step-by-step explanation:

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Answer 2

Answer:

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