Find the first five terms for the following arithmetic sequence Tn=5n-6?
1 answer:
Answer:
The first five terms are -1,4,9,14 and 19.
Step-by-step explanation:
The arithmetic sequence is :
We need to find the first five terms of the sequence.
Put n = 1 for first term.
Put n = 2 for second term.
Put n = 3 for third term.
Put n = 4 for fourth term.
Put n = 5 for fifth term.
Hence, the first five terms are -1,4,9,14 and 19.
You might be interested in
Hello,
a)
![I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\ = [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\ =0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\ ](https://tex.z-dn.net/?f=I%3D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5En%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%28x%29%2Asin%5E%7Bn-1%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D%20%5B-cos%28x%29%2Asin%5E%7Bn-1%7D%28x%29%5D_0%5E%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2B%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bcos%28x%29%2Asin%5E%7Bn-2%7D%28x%29%2Acos%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D0%20%2B%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bcos%5E2%28x%29%2Asin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7B%281-sin%5E2%28x%29%29%2Asin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20-%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5En%28x%29%20%5C%2C%20dx%5C%5C%0A%0A)
b)
![\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\ = \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\ = \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\ ](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5E%7B3%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%5Cfrac%7B2%7D%7B3%7D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%5Cdfrac%7B2%7D%7B3%7D%5C%20%5B-cos%28x%29%5D_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%3D%5Cdfrac%7B2%7D%7B3%7D%20%5C%5C%0A%0A%0A%0A%0A)
c)
a)
b)
c)