First term `a=10`, Common ratio `r=2`, Number of terms `n=10`
Find `n^(th)` term (last term) and sum of the geometric progression
Solution:
Here `a=10,r=2,n=10`
We know that,
`a_n = a × r^(n-1)`
`a_10=10×2^(10 - 1)`
`=10×512`
`=5120`
We know that,
`S_n = a * (r^n - 1)/(r - 1)`
`:.S_10=10× (2^10 - 1)/(2 - 1)`
`=>S_10=10× (1024 - 1)/1`
`=>S_10=10×1023/1`
`=>S_10=10230`
Hence, `10^(th)` term of the given series is `5120` and sum of first `10^(th)` term is `10230`
This material is intended as a summary. Use your textbook for detail explanation.
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