Sequences & Series Question - None

Question

In the sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ..., where n consecutive terms have the value n, find the 150th term of the sequence.

Accepted Answer

Step 1: Let the 150th term be n. Then the total count of terms up to value (n-1) should be less than 150, and the total count up to value n should be at least 150.

Step 2: The count of terms up to value (n-1) is $1 + 2 + 3 + \ldots + (n-1) = \frac{(n-1)n}{2}$

Step 3: The count of terms up to value n is $1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}$

Step 4: We need $\frac{(n-1)n}{2} < 150 \leq \frac{n(n+1)}{2}$

Step 5: From the first inequality: $n(n-1) < 300$ gives $n^2 - n - 300 < 0$, so $n < 17.8$ approximately.

Step 6: From the second inequality: $n(n+1) \geq 300$ gives $n^2 + n - 300 \geq 0$, so $n \geq 16.8$ approximately.

Step 7: Therefore, $16.8 < n < 17.8$, which means $n = 17$.

∴ The 150th term is 17.

Question:

Step-by-Step Solution

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