On the first day the bakery made 200 buns. Every other day the bakery made 5 buns more than the last day and this went up until the bakery made 1695 buns in one day. How many buns did the bakery make in total?

1 Answer
Oct 17, 2017

Rather long as I have not just jumped into the formula. I have explained the workings as I wish you to understand how the numbers behave.

44850200

Explanation:

This is the sum of a sequence.

First lets see if we can build an expression for the terms

Let i be the term count
Let a_i be the i^("th") term

a_i->a_1=200
a_i->a_2=200+5
a_i->a_3=200+5+5
a_i->a_4=200+5+5+5

On the last day we have 200+x=1695=>color(red)(x=1495)

and so on

By inspection we observe that as the general expression
for any color(white)(".")i we have a_i=200+5(i-1)

I am not going to algebraically solve this but the algebraic general term for the sum is:

sum_(i=1ton) [200+5(i-1)]

Instead lets try and reason this out.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let the sum be s

The actual sum numbers for n terms are:

s=200+(200+5)+(200+10)+(200+15)+....+[ 200+5(color(red)(1495)/5) ]

Note that 5((1495)/5) ->1495
This is the same as:

s=200+200[5+10+15+...+5(1495/5)]....Equation(1)

But the [5+10+15+....] is the same as
5[1+2+3+..+(n-1)]

So Equation(1) becomes

s=200+{200xx5[ color(white)(2/2) 1+2+3+5+...+(1495/5)color(white)(2/2)]color(white)(2/2)}

Factoring out the 200

s=200(1+5[color(white)(2/2)1+2+3+4+5+...+(1495/5)color(white)(2/2)]color(white)("d"))

s=200(1+5[color(white)(2/2)1+2+3+4+5+...+(299)color(white)(2/2)]color(white)("d"))

Notice that :

299+1=300
298+2=300
297+3=300

This is part of the process of determining the mean

So if we think on the lines of multiply the count of pairs by 300 we are on the way to determining the sum.

Consider the example: 1+2+3+4+5+6+7
The last number is odd and if we pair them up there is one value in the middle on its own. We don't want that!
So if we remove the first value we have an even count and thus all pairs. So remove 1 from [1+2+3+4+...+299] then we end up with:

299+2=301
298+3=301

So now we have n/2xx("first+last")->n/2xx(301)

The count n is 299-1=298 as we have removed the first number which is 1. So n/2->298/2 giving

1+298/2(2+299)color(white)("dddd")->color(white)("dddd")color(blue)(1+298xx(2+299)/2=44850)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Thus:

s=200(1+5[color(white)(2/2)1+2+3+4+5+...+(299)color(white)(2/2)]color(white)("d"))

becomes: color(red)(s=200(1+5(44850)) = 44850200)