First term `a=1`, Common ratio `r=3`, Number of terms `n=5`
Find `n^(th)` term (last term) and sum of the geometric progression
Solution:
Here `a=1,r=3,n=5`
We know that,
`a_n = a × r^(n-1)`
`a_5=1×3^(5 - 1)`
`=1×81`
`=81`
We know that,
`S_n = a * (r^n - 1)/(r - 1)`
`:.S_5=1× (3^5 - 1)/(3 - 1)`
`=>S_5=1× (243 - 1)/2`
`=>S_5=1×242/2`
`=>S_5=121`
Hence, `5^(th)` term of the given series is `81` and sum of first `5^(th)` term is `121`
This material is intended as a summary. Use your textbook for detail explanation.
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