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

35

Step-by-step explanation:

We know that the general formula for arithmetic sequence is  

a_n=a_0+(n-1)d

Using the general formula, we can make equations for 6th term and 9th term

a_6=a_0+(6-1)d

a_6=a_0+5d

101=a_0+5d Eqn 1

For 9th term

a_9=a_0+(9-1)d

a_9=a_0+8d

83=a_0+5d   Eqn 2

Subtracting equation 2 from equation 1

101-83=a_0+5d-(a_0+8d)

18=a_0+5d-a_0-8d  

18=-3d

d= -6

Putting d=-6 in Eqn 1, we get:

101=a_0+5(-6)

101=a_0-30

a_0=101+30

a_0=131

Now that we have values of a_0, and d we can find the 17th term

a_17=a_0+(17-1)d

=131+(16)(-6)

=131-96

=35

So the 17th term is 35 ..

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3