Answer:
see explanation
Step-by-step explanation:
The common difference d of an arithmetic sequence is
d =
-
=
- 
Substitute in values and solve for k, that is
5k - 1 - 2k = 6k + 2 - (5k - 1)
3k - 1 = 6k + 2 - 5k + 1
3k - 1 = k + 3 ( subtract k from both sides )
2k - 1 = 3 ( add 1 to both sides )
2k = 4 ⇒ k = 2
--------------------------------------------------------
The n th term of an arithmetic sequence is
=
+ (n - 1)d
= 2k = 2 × 2 = 4 and
d = 5k - 1 - 2k = 3k - 1 = (3 × 2) - 1 = 5
Hence
= 4 + (7 × 5) = 4 + 35 = 39
Answer:
f(x)=8*(3.5)^x
Step-by-step explanation:
2=98
3= 98r
4= 98r^2
5= 98r^3
6= 98r^4
98r^4/98, 98 crosses out and its just r^4. r^4= 14706.125/98
r=3.5
So you plug that in and you get f(x)=8*(3.5)^x
I hope this helped. XD

Let's solve ~




It's factorized now ~ And if you want to find the zeros then equate the expression with 0.
And you will get ;
