Find the 63rd term of the arithmetic sequence -1, 5, 11, ...−1,5,11,...
1 answer:
Answer:
![a_6_3=371](https://tex.z-dn.net/?f=a_6_3%3D371)
Step-by-step explanation:
we know that
In an <u><em>Arithmetic Sequence</em></u> the difference between one term and the next is a constant and this constant is called the common difference
we have
![-1,5,11,...](https://tex.z-dn.net/?f=-1%2C5%2C11%2C...)
Let
![a_1=-1\\a_2=5\\a_3=11](https://tex.z-dn.net/?f=a_1%3D-1%5C%5Ca_2%3D5%5C%5Ca_3%3D11)
![a_2-a_1=5-(-1)=5+1=6](https://tex.z-dn.net/?f=a_2-a_1%3D5-%28-1%29%3D5%2B1%3D6)
![a_3-a_2=11-5=6](https://tex.z-dn.net/?f=a_3-a_2%3D11-5%3D6)
The common difference is ![d=6](https://tex.z-dn.net/?f=d%3D6)
We can write an Arithmetic Sequence as a rule
![a_n=a_1+d(n-1)](https://tex.z-dn.net/?f=a_n%3Da_1%2Bd%28n-1%29)
where
a_n is the nth term
d is the common difference
a_1 is the first term
n is the number of terms
Find the 63rd term of the arithmetic sequence
we have
![n=63\\d=6\\a_1=-1](https://tex.z-dn.net/?f=n%3D63%5C%5Cd%3D6%5C%5Ca_1%3D-1)
substitute
![a_6_3=-1+6(63-1)](https://tex.z-dn.net/?f=a_6_3%3D-1%2B6%2863-1%29)
![a_6_3=-1+6(62)](https://tex.z-dn.net/?f=a_6_3%3D-1%2B6%2862%29)
![a_6_3=-1+372](https://tex.z-dn.net/?f=a_6_3%3D-1%2B372)
![a_6_3=371](https://tex.z-dn.net/?f=a_6_3%3D371)
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