1. First term `a=1`, Common ratio `r=2`, Number of terms `n=10`
Find `n^(th)` term (last term) and sum of the geometric progression

Solution:
Here `a=1,r=2,n=10`

We know that,
`a_n = a × r^(n-1)`

`a_10=1×2^(10 - 1)`

`=1×512`

`=512`

We know that,
`S_n = a * (r^n - 1)/(r - 1)`

`:.S_10=1× (2^10 - 1)/(2 - 1)`

`=>S_10=1× (1024 - 1)/1`

`=>S_10=1×1023/1`

`=>S_10=1023`

Hence, `10^(th)` term of the given series is `512` and sum of first `10^(th)` term is `1023`