The average of 7 consecutive odd numbers is M. If the next 3 odd numbe

The average of 7 consecutive odd numbers is M. If the next 3 odd numbers are also included, the average

remains unchanged

increases by 1.5

increases by 2

increases by 3

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC CAPF – 2017

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Let the 7 consecutive odd numbers be represented by an arithmetic progression with a common difference of 2. Let the middle term (4th term) be M, since for an odd number of terms in an AP, the average is the middle term. The 7 numbers are $M-6, M-4, M-2, M, M+2, M+4, M+6$. Their average is M.
The next 3 consecutive odd numbers after M+6 are $M+8, M+10, M+12$.
The new set of numbers consists of the original 7 plus these 3, totaling 10 numbers: $M-6, M-4, M-2, M, M+2, M+4, M+6, M+8, M+10, M+12$.
To find the new average, we sum these 10 numbers and divide by 10.
Sum of the first 7 numbers is $7M$.
Sum of the next 3 numbers is $(M+8) + (M+10) + (M+12) = 3M + 30$.
Total sum of the 10 numbers = $7M + (3M + 30) = 10M + 30$.
New average = $(10M + 30) / 10 = M + 3$.
The original average was M. The new average is M+3. The increase in average is $(M+3) – M = 3$.

When adding consecutive terms to an arithmetic progression, the average shifts. Adding terms that are all greater than the current average increases the average.

For any arithmetic progression, adding $k$ terms immediately following the last term of a sequence of $n$ terms will result in a new sequence of $n+k$ terms. The increase in average depends on $n$, $k$, and the common difference $d$. In this case, the common difference is 2. Adding $k=3$ terms after $n=7$ terms, the average increases by $(k \times d / 2) \times (n / (n+k))$ related term? No, simpler calculation method is better. The average of the $n$ numbers is $A_n$. The average of the $n+k$ numbers $A_{n+k}$ is $\frac{n A_n + \text{Sum of } k \text{ terms}}{n+k}$. The $k$ terms are $a_{n+1}, \ldots, a_{n+k}$. The average of the $n$ terms is $a_1 + (n-1)d/2$. The average of the $k$ new terms is $a_{n+1} + (k-1)d/2$. This method is more complex than the one used in the primary explanation using M as the middle term. The increase of 3 is consistent.

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