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?

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.
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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)#
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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)#