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kvv77 [185]
3 years ago
15

, list the first five terms of each sequence, and identify them as arithmetic or geometric.

Mathematics
1 answer:
sveta [45]3 years ago
5 0

Answer: -2, 2, 6, 10, 14

It is an Arithmetic sequence.

Step-by-step explanation:

Given Recursive formula : A(n + 1) = A(n) + 4 ,  for n ≥ 1 and A(1) = -2.

A(2)=A(1 + 1) = A(1) + 4=-2+4=2

A(3)=A(2 + 1) = A(2) + 4=2+4=6

A(4)=A(3 + 1) = A(3) + 4=6+4=10

A(5)=A(4 + 1) = A(4) + 4=10+4=14

Hence, the  five terms of given sequence= -2, 2, 6, 10, 14

Since, there is common difference of 4 in two consecutive terms.

Therefore, it is an Arithmetic sequence.

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PLEASE HELP!!!!! How many 4-digit numbers divisible by 5, all of the digits of which are even, are there?
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Answer:

I assume you know Arithmetic Progression .

so, we have to find the first and last 4-digit number divisible by 5

first = 1000 ,  last =  9990

we have a formula,   a_{n} = a + (n-1)d

here, a_{n} is the last 4-digit number divisible by 5.

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3 years ago
What is the 6th term of the geometric sequence where a1 = -4096 and a4 = 64?
Akimi4 [234]
\bf \begin{array}{llccll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_1&-4096\\
a_2&-4096r\\
a_3&-4096rr\\
a_4&-4096rrr\\
&-4096r^3\\
&64
\end{array}\implies -4096r^3=64
\\\\\\
r^3=\cfrac{64}{-4096}\implies r^3=-\cfrac{1}{64}\implies r=\sqrt[3]{-\cfrac{1}{64}}
\\\\\\
r=\cfrac{\sqrt[3]{-1}}{\sqrt[3]{64}}\implies \boxed{r=\cfrac{-1}{4}}\\\\
-------------------------------

\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
r=-\frac{1}{4}\\
a_1=-4096\\
n=6
\end{cases}
\\\\\\
a_6=-4096\left( -\frac{1}{4} \right)^{6-1}\implies a_6=-4^6\left( -\frac{1}{4} \right)^5
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