Average
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Average & the sum bridge
Definition
\bar{x} = \frac{\sum x_i}{n} \iff \sum x_i = n\bar{x}
The sum form is the one you use.
Shift property
\overline{x_i + k} = \bar{x} + k,\quad \overline{k\, x_i} = k\bar{x}
Adding k shifts the average by k; multiplying by k scales it.
First n naturals / odd / even
\frac{n+1}{2},\quad n,\quad n+1
n = how many terms.
Squares / cubes
\frac{(n+1)(2n+1)}{6},\quad \frac{n(n+1)^2}{4}
Equal gaps
\text{average} = \frac{\text{first} + \text{last}}{2} = \text{middle term}
Members joining or leaving
Joining member
\text{value} = A' + n(A' - A)
n = old count; A' = new average.
Leaving member
\text{value} = A' + n(A - A')
A' = average of the remaining n members.
Replacement
\text{new} = \text{old} + n(A' - A)
The count does not change.
Count change both ways
\text{total of newcomers} = \text{new total} - \text{old total}
Weighted average & two-group problems
Weighted mean
\bar{x} = \frac{\sum n_i \bar{x}_i}{\sum n_i}
Missing group average
\bar{x}_2 = \frac{N\bar{x} - n_1\bar{x}_1}{n_2}
N = total count, overall average known.
Sizes from distances
\frac{n_1}{n_2} = \frac{\bar{x}_2 - \bar{x}}{\bar{x} - \bar{x}_1}
Reverse ratio of the distances.
Batsman problems, overlapping sums & multi-step sets
Batsman score
\text{score} = A' + (n-1)d
A' = new average, d = rise.
Batsman new average
A' = \frac{(n-1)A + \text{score}}{n} = x - (n-1)d
Overlap subtraction
\text{Thu} - \text{Mon} = 3(b - a)
Three-day windows; multiply by the overlap length.
Shared middle item
\text{shared} = S_1 + S_2 - S_{\text{total}}
Split sums
n\bar{x} = \sum_{\text{chunks}} (\text{chunk total})
One equation per chunk.