These are the first six terms of a sequence with a = 2:
2, 14, 98, 686, 4802, 33614,
find a recursive formula for this sequence that is valid for n > 1.
write your answer in simplest form.

Step-by-step explanation:

98÷14

=7

686÷98

=7

4 802÷686

=7

33 614÷4 802

=7

Then the common ratio q for this sequence is 7

recursive formula : an+1 = q×an = ?

an= a1 × qⁿ⁻¹

   =2×7ⁿ⁻¹

an+1 = q×an

        = 7×(2×7ⁿ⁻¹)

       = 2×7ⁿ

The formular for this sequence is AR^n-1 (that is, A multiplied by R{raised to the power of n minus 1} )

Step-by-step explanation:This is a geometric progression in which every term is calculated by multiplying each previous term by a common ratio.

The common ratio here is 7, which is derived as

14/2, or 98/14, or 686/98, or 4802/686...

In simply put, R is derived as Tn/Tn-1, where Tn is the nth term and Tn-1 is the previous term.

Therefore the formular for this progression is given as

AR^n-1

Where A = 2, R = 7 and n = the nth term.

y=x+9

?

step-by-step explanation:

something over 10 so ? /10

step-by-step explanation: