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.

1,589.63 in.^3

step-by-step explanation:

i hope this you : d

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it