Question

The average of a non-decreasing sequence of N numbers \(a_1,a_2,…,a_N\) is 300.If \(a_1\) is replaced by \(6a_1\), the new average becomes 400.Then,the number of possible values of \(a_1\) is

Answer 1

The correct answer is: 14 

Given:
Average of the non-decreasing sequence of N numbers, \(a_1, a_2, ..., a_N = 300\)

When \(a_1\) is replaced by \(6a_1\), the new average becomes 400.

We know:
Average = \(\frac{\text{Sum of elements}}{N}\)

Original sum = \(300N\)
New sum = \(400N\)

Since only \(a_1\) is replaced by \(6a_1\),
New sum = \(300N - a_1 + 6a_1 = 300N + 5a_1\)

Equating to the new sum:
\(300N + 5a_1 = 400N\)
\(5a_1 = 100N\)
\(a_1 = 20N\)

Since \(a_1\) must be a positive integer in a non-decreasing sequence, N must also be a positive integer.
We also know from the original average that:
\(\frac{a_1 + a_2 + \cdots + a_N}{N} = 300 \Rightarrow a_1 + a_2 + \cdots + a_N = 300N\)

But \(a_1 = 20N\), so it must be less than or equal to 300:
\(20N \leq 300 \Rightarrow N \leq 15\)

Also, since a sequence with only one element can't be "non-decreasing" meaningfully, \(N \geq 2\)

So possible values of N: 2 to 15 (inclusive), i.e., 14 values
Each gives a unique \(a_1 = 20N\): 40, 60, 80, ..., 300

∴ The number of possible values of \(a_1\) is 14.

Answer 2

Given: 
\(a_1 + a_2 + \dots + a_N = 300N \quad \text{(1)}\)
\(6a_1 + a_2 + \dots + a_N = 400N \quad \text{(2)}\)

Step 1: Subtract equation (1) from equation (2)
\((6a_1 + a_2 + \dots + a_N) - (a_1 + a_2 + \dots + a_N) = 400N - 300N\)
\(5a_1 = 100N\)
\(a_1 = \frac{100N}{5} = 20N\)

Step 2: Use equation (1) to find the average
Total sum = \(a_1 + a_2 + \dots + a_N = 300N\)
So average = \(\frac{300N}{N} = 300\)

Step 3: From above, \(a_1 = 20N\)
But since the sequence is non-decreasing, all other terms must be ≥ \(a_1\)
Therefore, \(a_1 \leq 300\)

Now substitute: \(20N \leq 300 \Rightarrow N \leq 15\)
Also, \(N \ne 1\) because then all terms would be \(a_1 = 20 \cdot 1 = 20\), but sum would be 20 not 300.

Hence:
\(2 \leq N \leq 15\)
So, number of possible values = \(15 - 2 + 1 = 14\)

Final Answer: \(14\) possible values of \(N\)