Question 1164947
It is a series with the first term 6 m out and back or 12 m
the second term is 11 m out and back or 22 m
and so forth until the n th term is (6+5(n-1)) out and back or 12+10n-10 m or 10n +2
The last term will be 102 m with n=10

so the series is 12+22+32+...+102

a1=12
a10=102
sum is (n/2)(a1+an)
=5(12+102)
=570 m

can check 
12+22+32+42+52+62+72+82+92+102