Arithmetic Series (defined limits)

If the first term of an arithmetic progression is $t_1=a$t1​=a and the common difference is $d$d , then the sequence becomes:        

  $t_1=a$t1​=a ,  $t_2=a+d$t2​=a+d ,  $t_3=a+2d$t3​=a+2d , $t_4=a+3d$t4​=a+3d , $t_5=a+4d$t5​=a+4d,  ...           

Suppose we continue writing the terms all the way up to $t_{100}=a+99d$t100​=a+99d , and then find a way to add up all of the one hundred terms together. The great German mathematician Carl Friedrich Gauss in 1785, at the age of 8 years old, used the following method.

Instead of adding the terms from left to right, he added them in this order:

$\left(t_1+t_{100}\right)+\left(t_2+t_{99}\right)+\left(t_3+t_{98}\right)+$(t1​+t100​)+(t2​+t99​)+(t3​+t98​)+ $...+\left(t_{48}+t_{52}\right)+\left(t_{49}+t_{51}\right)$...+(t48​+t52​)+(t49​+t51​)

Can you see why he may have done that?  

Every bracketed pair of terms adds up to $2a+99d$2a+99d and since there are fifty pairs, their total must be $50\times\left(2a+99d\right)$50×(2a+99d). Gauss knew there were $50$50 pairs because there are $100$100 terms. His answer could be written $\frac{100}{2}\times\left(2a+99d\right)$1002​×(2a+99d) .

This led him to consider a formula for adding any number of terms of any arithmetic progression. Writing $S_n$Sn​ for the sum to $n$n terms, Gauss was able to show that:

$S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$Sn​=n2​[2a+(n−1)d] 

For example, adding all of the positive integers from $1$1 to $49$49 becomes an easy task to perform. With $a=1$a=1  and $d=1$d=1 we have: 

$S_{49}=\frac{49}{2}\left[2\times1+\left(49-1\right)\times1\right]=1225$S49​=492​[2×1+(49−1)×1]=1225

Note that the formula works even though there are an odd number of terms to add up.  Can you explain why?

If we know the last term  of the arithmetic progression, then the task becomes even easier. Splitting the $a$a's up in the formula we can write $S_n$Sn​ as $S_n=\frac{n}{2}\left[a+a+\left(n-1\right)d\right]$Sn​=n2​[a+a+(n−1)d] and since the last term $l=t_n=a+\left(n-1\right)d$l=tn​=a+(n−1)d , deduce that:

$S_n=\frac{n}{2}\left[a+l\right]$Sn​=n2​[a+l]

For example, to find the sum of the fourteen multiples of $7$7 between $100$100 and $200$200, we first discover, by dividing, that $a=105$a=105 and $l=196$l=196 and thus:

$S_{14}=\frac{14}{2}\left[105+196\right]=2107$S14​=142​[105+196]=2107

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3