The average of 5 consecutive number integers starting with m as the first integers is n then n = ?

The average of 5 consecutive number integers starting with m as the first integers is n then n = ? ?

Created: 1 year ago | Updated: 2 months ago | Created By:

Jannatul Ferdush

m+3

m+2

nm+2

0 20

Institution: Bangladesh Bank Exam: সমন্বিত ৪ ব্যাংক - সহকারী প্রোগ্রামার - 25.09.2020 ||
Topic: Arithmetic Mathematics (পাটিগণিত), Subject: Mathematics (সাধারণ গণিত)

Best Answer Explained in Below-:

To understand why the average of 5 consecutive integers starting from \( m \) is \( m + 2 \), let’s first identify the 5 consecutive integers. They can be represented as \( m, m + 1, m + 2, m + 3, \) and \( m + 4 \).

To find the average of these integers, we sum them up:

\[
m + (m + 1) + (m + 2) + (m + 3) + (m + 4) = 5m + (0 + 1 + 2 + 3 + 4) = 5m + 10
\]

Next, we divide this total by the number of integers, which is 5:

\[
\text{Average} = \frac{5m + 10}{5}
\]

Simplifying this expression:

\[
\text{Average} = m + \frac{10}{5} = m + 2
\]

Thus, the average of the 5 consecutive integers starting with \( m \) is indeed \( m + 2 \). This aligns with option C from the provided choices, confirming that the correct answer is \( C: m + 2 \). Options A, B, and D do not correctly represent the average of these integers based on the calculations.

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The average of 5 consecutive number integers starting with m as the first integers is n then n = ? ?

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