Consider the sequence 1, – 2, 3, – 4, 5, – 6, . . . . . . \[n{( - 1)^{n + 1}}\]. What is the average of the first 300 terms of the sequence?
(a) – 1
(b) 0.5
(c) 0
(d) – 0.5

Hint: Separate the terms so that they form two arithmetic progressions (A.P.s), then find the sum of both A.P.s and add the sum and divide by 300 to get the average of the first 300 terms of the sequence.

Complete step-by-step answer:
We need to find the average of the first 300 terms of the sequence, 1, – 2, 3, – 4, 5, – 6, . . . . . . \[n{( - 1)^{n + 1}}\].

The first 300 terms of this sequence are as follows:

1, – 2, 3, – 4, 5, – 6, . . . . . ., 299, – 300

We can write this sequence as two separate arithmetic progressions (A.P.s) as follows:

1, 3, 5, . . . , 299 and – 2, – 4, – 6, . . . , – 300

The sum of n terms of an A.P. with first term a and last term l is given as follows:

\[S = \dfrac{n}{2}(a + l)...........(1)\]

Let us consider the A.P.: 1, 3, 5, . . . , 299 , we have:

\[a = 1\]

\[l = 299\]

\[n = 150\]

Using the formula (1) to calculate the sum of this A.P., we have:

\[{S_1} = \dfrac{{150}}{2}(1 + 299)\]

\[{S_1} = \dfrac{{150}}{2}(300)\]

\[{S_1} = 150 \times 150\]

\[{S_1} = 22500...........(2)\]

Next, we consider the A.P.: – 2, – 4, – 6, . . . , – 300, we have:

\[a = - 2\]

\[l = - 300\]

\[n = 150\]

Using the formula (1) to calculate the sum of this A.P., we have:

\[{S_2} = \dfrac{{150}}{2}( - 2 - 300)\]

\[{S_2} = \dfrac{{150}}{2}( - 302)\]

\[{S_2} = 150( - 151)\]

\[{S_2} = - 22650.........(3)\]

Adding equations (2) and equation (3), we have:

\[{S_1} + {S_2} = 22500 - 22650\]

\[{S_1} + {S_2} = - 150\]

The average of the first 300 terms is the sum of the 300 terms divided by 300.

Average = \[\dfrac{{ - 150}}{{300}}\]

Average = – 0.5

Hence, the correct answer is option (d).

Note: Take the negative sign into account when you do your calculations. sign into account when you do your calculations. You can also add two nearby terms and the answer is – 1, and conclude that the average is – 1 divided by 2, that is, – 0.5.