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`
This material is intended as a summary. Use your textbook for detail explanation.
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