If sum of n terms of an AP is 5n2-3n, then sum of its 100th term is

1.	If sum of n terms of an AP is 5n2-3n, then sum of its 100th term is
Answer» \(\sum_{i=1}^n \) a_i = 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) i^th term of sequence = 10i - 8 100^th term of sequence = 1000-8 = 992 Sum of digits = 9 + 9 + 2 = 20

Answer»

\(\sum_{i=1}^n \) a_i = 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)

i^th term of sequence = 10i - 8

100^th term of sequence = 1000-8 = 992

Sum of digits = 9 + 9 + 2 = 20

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