Question

How do you find `S_n` for the arithmetic series given `a_1=-64`, d=-3, n=21?

Answer 1

`-1974`

Explanation:

You could just use the formula, plug in the values and do the calculation. This was referred to by one of my recent contact as 'plug and chug'

This will not help you understand what is going on. The standardised formula is `S_n=1/2 n(2a_1+d[n-1])`.. Equation(1)
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`color(blue)("Building the formula")`

`S_n=-(64+67+70+73+...+a_n)`

`S_n=-(64+(64+3)+(64+6)+(64+9)+...`

`S_n=-(64n+3+6+9+12+...`

`S_n=-(64n+3(1+2+3+4+...+n-1)`
.............................................................................
Note that `sum(1+2+3+...+(n-1))`

`= ("first term+last term")/2xx"count" -> (1+(n-1))/2xx(n-1)`

`=n/2(n-1)`

However we have `3sum(1+2+...`

So we write `(3n)/2(n-1)`
.............................................................................

So `S_n=-[" "64n+(3n)/2(n-1)" "]` .................Equation(2)

Given that `n= 21` we have

`S_n=-[" "64(21)+63/2(21-1)" "]`

Giving:

`" "color(blue)(bar(ul(|color(white)(2/2)S_n=-(1344+630) = -1974" "|)))`
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`color(blue)("Consider equation(2) but as")`

`S_n=64n+(3n)/2(n-1)`

but 64 is the first term so we have

`S_n=a_1n+(3n)/2(n-1)`

Factor out `n/2`

`S_n=1/2 n[" "2a_1+3(n-1)" " ]`

But the 3 is the difference `d` so we write:

`color(brown)(S_n=1/2 n[" "2a_1+d(n-1)" " ] )`

`color(brown)("The above is the same as Equation(1)")`