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The 342nd term of the series 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, ……….. is

Question

The 342nd term of the series 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, ……….. is

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Solution

The pattern of the series is such that each number n is repeated n times.

To find the 342nd term, we can use the formula for the sum of the first n natural numbers, which is n*(n+1)/2.

We need to find a number n such that n*(n+1)/2 is just less than or equal to 342.

If we try n=26, we get 26*(26+1)/2 = 351, which is greater than 342.

If we try n=25, we get 25*(25+1)/2 = 325, which is less than 342.

So, the 342nd term is in the 26th set of repeated numbers.

To find the exact position within this set, subtract the total number of terms up to the 25th set from 342: 342 - 325 = 17.

So, the 342nd term is the 17th term in the 26th set of repeated numbers.

Since each set contains the same number repeated, the 342nd term of the series is 26.

This problem has been solved

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