\(\sum_{i=1}^n \) ai = 5n2 - 3n
= 5n2 + 5n - 8n
= 10\(\big(\frac{n^2+n}{2}\big)\)- 8n
=10 \(\sum_{i=1}^n \) i - 8 \(\sum_{i=1}^n \)1
= \(\sum_{i=1}^n \)(10 i -8)
ith term of sequence = 10i - 8
100th term of sequence = 1000-8 = 992
Sum of digits = 9 + 9 + 2 = 20