Question

PLEASE HELP.

Answer 1

Answer:

Explicit formula for

1) $a_n=17-3(n-1)$

2) $a_n=(\frac{1}{2} )^{n-1}\cdot 20$

Step-by-step explanation:

W have to find the first few terms of the given sequence and then find an explicit equation

Explicit formula = f(n) + d(n-1) , where,

f(n) is first term,

d is common difference,

n-1 is one term less than the term number.

1)

Given : $a_1=17\\\\a_{n+1}=a_n-3$

$\text{We put n =1, we get,}\\\\a_2=a_1-3=17-3=11=17-3(2-1)\\\\\\\text{We put n =2, we get,}\\\\\\a_3=a_2-3=11-3=8=17-3-3=17-3\cdot 2=17-3(3-1)\\\\\text{We put n =3, we get,}\\\\\\a_4=a_3-3=8-3=5=17-3-3-3=17-3\cdot 3=17-3(4-1)$

Thus, we obtained an explicit formula,

$a_n=17-3(n-1)$

2)

Given : $a_1=20\\\\a_{n+1}=\frac{1}{2} \cdot a_n$

$\text{We put n =1, we get,}\\\\a_2=a_1-3=17-3=11=17-3(2-1)\\\\\\\text{We put n =2, we get,}\\\\\\a_3=a_2-3=11-3=8=17-3-3=17-3\cdot 2=17-3(3-1)\\\\\text{We put n =3, we get,}\\\\\\a_4=a_3-3=8-3=5=17-3-3-3=17-3\cdot 3=17-3(4-1)$

Thus, we obtained an explicit formula,

$a_n=17-3(n-1)$

$\text{We put n =1, we get,}\\\\\\a_2=\frac{1}{2} \cdot a_1=\frac{1}{2} \cdot 20=10=(\frac{1}{2} )^{2-1}\cdot 20\\\\\\\text{We put n =3, we get,}\\\\\\a_4=\frac{1}{2} \cdot a_3=\frac{1}{2} \cdot 5=\frac{5}{2}=(\frac{1}{2} )^{4-1}\cdot 20$

Thus, we obtained an explicit formula,

$a_n=(\frac{1}{2} )^{n-1}\cdot 20$

Answer 2

Answer:

1. Terms are 17, 14, 11, 8, 5,...... and explicit equation is $a_{n}=17-3(n-1)$.

2. Terms are 20, 10, 5, 2.5, 1.25,...... and explicit equation is $a_{n}=20\times (\frac{1}{2})^{n-1}$.

Step-by-step explanation:

Ques 1: We are given the recursive formula for the sequence as,

$a_{n+1}=a_{n}-3$, where $a_{1}=17$.

So, substituting the values of 'n' from {1,2,3,.....}, we get,

$a_{2}=a_{1+1}=a_{1}-3=17-3=14$

$a_{3}=a_{2+1}=a_{2}-3=14-3=11$

$a_{4}=a_{3+1}=a_{3}-3=11-3=8$

$a_{5}=a_{4+1}=a_{4}-3=8-3=5$

Thus, the sequence is given by 17, 14, 11, 8, 5,......

As, the explicit equation of an arithmetic sequence is of the form, $a_{n}=a_{1}+d(n-1)$, where $a_{1}$ is the first term and 'd' is the common difference.

As, the common difference, d = 14 - 17 = -3

Thus, we get,

The given sequence has the explicit equation, $a_{n}=17-3(n-1)$.

Ques 2: We are given the recursive formula for the sequence as,

$a_{n+1}=\frac{a_{n}}{2}$, where $a_{1}=20$.

So, substituting the values of 'n' from {1,2,3,.....}, we get,

$a_{2}=a_{1+1}=\frac{a_{1}}{2}=\frac{20}{2}=10$

$a_{3}=a_{2+1}=\frac{a_{2}}{2}=\frac{10}{2}=5$

$a_{4}=a_{3+1}=\frac{a_{3}}{2}=\frac{5}{2}=2.5$

$a_{5}=a_{4+1}=\frac{a_{4}}{2}=\frac{2.5}{2}=1.25$

Thus, the sequence is given by 20, 10, 5, 2.5, 1.25,......

As, the explicit equation of a geometric sequence is of the form, $a_{n}=a_{1}\times r^(n-1)$, where $a_{1}$ is the first term and 'r' is the common ratio.

As, the common ratio, $r=\frac{10}{20}=\frac{1}{2}$

Thus, we get,

The given sequence has the explicit equation, $a_{n}=20\times (\frac{1}{2})^{n-1}$.