Question:
For any natural number n, suppose the sum of the first n terms of an arithmetic progression is n(n+1). If the nth term of the progression is divisible by 7, then the smallest possible value of n is:
For any natural number n, suppose the sum of the first n terms of an arithmetic progression is n(n+1). If the nth term of the progression is divisible by 7, then the smallest possible value of n is:
Updated On: Sep 17, 2024
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The Correct Option is B
Solution and Explanation
We know, nth term of an AP = Sum of n terms - Sum of (n-1) terms
Therefore, nth term = \(n(n+1) - (n-1)(n) = n^2 + n - n^2 + n = 2n\)
Given, nth term is divisible by 7. So, 2n must be divisible by 7.
For 2n to be divisible by 7, the smallest possible value of n is when \(2n = 7\times2 = 14.\)
Therefore, n = 7.
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