Question 344995
<pre>
We need to find their sum and the number of terms

The sum is

S<sub>n</sub> = 2+12+22+32+42+52+62+72+82+92+102+ ... +972+982+992

This is an arithmetic series with {{{a[1]=2}}} and {{{d=10}}} and {{{a[n]=992}}}

{{{a[n]=a[1]+(n-1)d}}}

{{{992 = 2 + (n-1)10}}}

{{{992=2+10(n-1)}}}

{{{992=2+10n-10}}}

{{{992=10n-8}}}

{{{1000=10n}}}

{{{100=n}}}

So there are 100 terms:

The sum formula is:

S<sub>n</sub> = {{{expr(n/2)*(a[1]+a[n])}}}

S<sub>100</sub> = {{{expr(100/2)*(2+992)}}}

S<sub>100</sub> = {{{50*(994)=49700}}}

So the average term is their sum divided by 100, the number of terms

{{{49700/100}}} = 497.

Edwin</pre>