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Chapter 18: Problem 702

The mean of the series \(a, a+d, a+2 d \ldots \ldots . a+(2 n+1) d\) is (a) \(a+[(2 n+1) / 2] d\) (b) \(a+(n+1) d\) (c) \(a+(2 n+1) d\) (d) \(a+[(2 n-1) / 2] d\)

Short Answer

Expert verified
The mean of the given series is \(a + \frac{(2n+1)d}{2}\).

Step by step solution

01

Calculate the sum of the series

We have the series: \(a, a+d, a+2d, \ldots, a+(2n+1)d\). We can see that this is an arithmetic series with common difference d. The sum of an arithmetic series can be calculated using the formula: \[S = \frac{n}{2} (a_1 + a_n)\] where n is the number of terms, \(a_1\) is the first term and \(a_n\) is the last term. In our case, \(a_1 = a\) and \(a_n = a + (2n+1)d\). The total number of terms in the series is (2n+2). Now, let's find the sum of the series, S: \[S = \frac{2n+2}{2} (a + a + (2n+1)d)\]
02

Simplify the sum formula

Upon simplifying the formula for the sum, we get: \[S = (n+1)(2a+2n d + d)\] \[S = (n+1)(2a+(2n+1)d)\]
03

Calculate the mean of the series

Now we will find the mean of the series, which is the sum of the series divided by the total number of terms. As mentioned earlier, the total number of terms is (2n+2). Mean = \(\frac{S}{2n+2}\) Substitute the formula for S from Step 2: \[Mean = \frac{(n+1)(2a+(2n+1)d)}{2n+2}\]
04

Simplify the mean formula

Simplify the mean formula: \[Mean = \frac{(n+1)(2a+(2n+1)d)}{(n+1)(2)}\] \[Mean = \frac{2a+(2n+1)d}{2}\] Our final expression for the mean is: \[Mean = a + \frac{(2n+1)d}{2}\] Now we can see that this matches option (a): \(a + \frac{(2n+1)d}{2}\). So, the correct answer is (a).

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Most popular questions from this chapter

The average of \(n\) numbers \(y_{1}, y_{2} \ldots \ldots . . y_{n}\) is M. If \(y_{n}\) is replaced by \(\mathrm{y}^{\prime}\) then the new average is (a) \(\left[\left(M+y_{n}-y^{\prime}\right) / n\right]\) (b) \(\left[\left\\{(n-1) M+y^{\prime}\right\\} / n\right]\) (c) \(\left[\left(n M-y_{n}+y^{\prime}\right) / n\right]\) (d) \(M-y_{n}-y^{\prime}\)

A dice is loaded so that the probability of face \(\mathrm{i}\) is proportional to i. \(\mathrm{i}=1,2, \ldots .6\). Then the probability of an even number occupy when the dice is rolled is (a) \((2 / 7)\) (b) \((3 / 7)\) (c) \((4 / 7)\) (d) \((5 / 7)\)

Standard deviation of \(-1,-2,-3,-4,-5,-6,-7\) is (a) \(-4\) (b) 4 (c) 2 (d) \(-2\)

Mean of n observations is \(\mathrm{m}\) and sum of \(\mathrm{n}-3\) observations is b then mean of remaining 3 observations is (a) \(n m+b\) (b) \([(\mathrm{nm}-\mathrm{b}) / 3]\) (c) \([(n m+b) / 3]\) (d) \(n m-b\)

Out of 3 n consecutive integers three are selected at random the probability that there sum is divisible by 3 is (a) \(\left[\left(3 n^{2}-n-2\right) /\\{(3 n-1)(3 n-2)\\}\right]\) (b) \(\left[\left(n^{2}-3 n+2\right) /\\{(3 n-1)(3 n-2)\\}\right]\) (c) \(\left[\left(3 n^{2}-3 n+2\right) /\\{(3 n-2)(3 n-3)\\}\right]\) (d) \(\left[\left(3 n^{2}-3 n+2\right) /\\{(3 n-1)(3 n-2)\\}\right]\)
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