1. First term `a=1`, Common difference `d=1`, Number of terms `n=100`
Find `n^(th)` term (last term) and sum of the arithmetic progression

Solution:
Here first term `a=1,`

Common difference `d=1`

We know that,
`f(n) = a + (n - 1)d`

`f(100)=1 + (100 - 1)(1)`

`=1 + (99)`

`=100`

We know that,
`S_n = n/2 [2a + (n - 1)d]`

`:. S_100 = 100/2 * [2(1) + (100 - 1)(1)]`

`= 50 * [2 + (99)]`

`= 50 * [101]`

`= 5050`

Hence, `100^(th)` term of the given series is `100` and sum of first `100^(th)` term is `5050`