Question

the sum of 18 consecutive natural number is a perfect square. what is the smallest possible value of this sum. ​

Answer 1

Answer:

Let the number be

a, a + 1, a + 2, a + 3 _ _ _ _ _ _ a + 17.

This is an AP.

First term = a

Common difference, d = 1

Terms, n = 18

Sum of the terms

Substitute the values :

Sn = n/2 [ 2a + (n - 1)d ]

=> Sn = 18/2 [ 2a + 17 ]

=> Sn = 9 [ 2a + 17 ]

=> Sn = 18a + 153

For a = 1, Sn = 171

a = 2, Sn = 189

a = 3, Sn = 207

a = 4, Sn = 225

Here, 225 is the perfect square of 15.

Thus, 15 is a perfect square the smallest possible value of this sum is 225.

Answer 2

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Sum of an Arithmetic Progression:

Sum = N * (First + Last )/2 =

= 18 * (a + a+17)/2 = 9*(2a+17)= 18a + 153

Sum = 9*(2a+17)

So, (2a+17) has to be a square… and that means an odd number greater than 17 that is a square: 19, 21, 23, 25.

(of course there are infinite odd numbers which are squares, but the smallest greater than 17 is 25)

9*25 = 225

225 is the smallest possible value of that sum

25 -17 = 8

Then a = 4

4+5+6+….+(4+17) = 4+5+6+…+19+20+21 =

= (21+4) + (20+5) + … +(12+13) =

= (25)*9 = 225