Question

A student was told to add first ‘n' natural numbers and find the sum. He got the sum as 1850. But when his teacher checked the sum, he found that one of the numbers was added twice and hence was the sum. Hence as the punishment, the teacher asked him to find the difference between ‘n' and that number. What was the difference?

Answer 1

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Answer 2

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$\mathbb{\underline{\green{SOLUTION:}}}$

¶ The sum of first n natural numbers:
$S = \frac{n(n+1)}{2}$

Natural No.s from 1 to n are added but 'n' is unknown.

However, the sum of 1 to n natural numbers has been given, i .e., 1850

$\frac{n(n+1)}{2} = 1850$

Also Given, this sum is wrong because the student had added one number(say k) twice.

Now find k which will not only enable us to find the value of n but also correct the sum.

=> n(n+1) = 1850 × 2

=> n² + n = 3700

By trial and Error method,
Choose arbitrary value of n satisfying the above equation

we know,
55 × 55 = (5×(5+1))|(5×5) = (5×6)|(5×5) = 3025
3025 < 3700

$57^{2} + 57 = 3306$

3306 < 3700
Take successor of 57 i. e., 58
$58^{2} + 58 = 3422$

3422 < 3700
Take successor of 58 i. e., 59
$59^{2} + 59 = 3540$

3540 < 3700
Take successor of 59 i. e., 60
$60^{2} + 60 = 3660$

(61 × 61) + 61 = 3782

If we take successor of 60, sum increases beyond 3700

Since twice added number will only increase the average, we will take n as 60.

•°• n = 60

3700 - 3660 = 40

$\frac{40}{2} = 20$

•°• k = 20

Now,

Given,
Faulty sum = 1850

¶ (Faulty sum)1850 = (correct sum) + Excess Value

¶ Excess Value = Faulty sum - Correct Sum

• Correct Sum =>
$\frac{60(60+1)}{2} = 1830$

• 1830 =1850- Excess Value
=> Excess value = 1850 - 1830

• Faulty sum=>
$\frac{60(60+1)}{2} + 20 = 1850$

so,
A possibility is k = 20 & N = 60
The difference is 60-20 = 40

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$\mathcal{\huge{\purple{Hope\: it\: helps}}}$