What is the 17th term in the arithmetic sequence in which a6 is 101 and a9 is 83

The 17th term in arithmetic sequence is 68

Step-by-step explanation:

The general formula of arithmetic sequence is:

aₙ = a₁ + (n – 1)d.

We are given a₆ = 101 and a₉ = 83 and we need to find a₁₇

To find the term a₁₇ we should know a₁ and d. So we would find both

a₆ = a₁ +(6-1)d

101 = a₁ +(5)d

101 = a₁ +5d     eq(1)

and

a₉ = a₁ +(9-1)d    

83 = a₁ + 8d       eq(2)

Subtracting eq(2) from eq(1)

101 = a₁ +5d

83 = a₁ + 8d

-       -     -

18 = -3d

=> d = 18/-3

=> d = -6

Putting value of d in eq(1)

101 = a₁ + 5d

101 = a₁ + 5(-3)

101 = a₁ -15

=> a₁ = 101+15

=> a₁ = 116

Now finding a₁₇:

aₙ = a₁ + (n – 1)d.

a₁₇ = 116 +(17-1)(-3)

a₁₇ = 116+(16)(-3)

a₁₇ = 116 - 48

a₁₇ = 68

So, the 17th term in arithmetic sequence is 68

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

im here

step-by-step explanation: