Question follows the 5th term of an exponential sequence is greater than the 4th term by thirteen and half and the 4th term is greater than the 3rd term by 9 .find the common ratio and the 1st term

I figured it out but was thrown off by the poor wording used. Please always copy the question exactly how it is given to you otherwise you might not get any help or wrong answer.

terms of geometric sequence
A(n) = ar^(n-1)
where "a" represents first term
here you are given information to set up two equations in two variables "a" and "r". you must combine equations to solve for one variable at a time.

A(4) - A(3) = 9
(ar^3) - ar^2 = 9
a(r^3/2 - r^2) = 9
a = 9/(r^3-r^2)

A(5) - A(4) = 13.5
ar^4 - ar^3 = 13.5
a(r^4 - r^3) = 13.5

combined
9/(r^3 - r^2) * (r^4 - r^3) = 13.5
cross multiply
9r^4 - 9r^3 = 13.5r^3 - 13.5r^2
combine like terms
9r^4 - 22.5r^3 + 13.5r^2 = 0
factor/polysolver
r = 3/2 or r = 1
but r cannot equal 1

ratio is 3/2 = 1.5

a = 9/(1.5^3 - 1.5^2) = 8
first term is 8

yes

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

For range i think it’s area