your first sequence looks geometric with
a = 81, r = 1/3
term(15) = ar^14
= 81(1/4782969)
= 1/59049
2. let the missing term be x
then x/39 = 975/x
x^2 = 38025
x = ± √38025 = ± 195
could be 195 or -195
3. looks arithmetic with a = -5 , d = 5
term(n) = 65
-5 + (n-1)(5) = 65
5n - 5 = 70
5n = 75
so we have 15 terms
sum(15) = (15/2)(first + last)
= 15/2(-5+65) = 450
I have 3 questions I need help with.
1) what is the 15th term of the sequence 81,27,9,...
2) what is a possible value for the missing term of the geometric sequence 39,_,975,_
3) what is the sum of the finite arithmetic series (-5)+0+5+10+..+65
3 answers
the first and second is not a possible answer.
1) 1/729
1/243
1/81
1/810
1) 1/729
1/243
1/81
1/810
1. proof that my answer is correct
81 27 9 3 1 1/3 1/9 1/27 1/81 1/243 1/729 1/2187 1/6561 1/19683 1/59049
count the terms
2. proof my answer is correct:
195/39 = 5
975/195 = 5
-195/39 - -5
975/-195 = -5
81 27 9 3 1 1/3 1/9 1/27 1/81 1/243 1/729 1/2187 1/6561 1/19683 1/59049
count the terms
2. proof my answer is correct:
195/39 = 5
975/195 = 5
-195/39 - -5
975/-195 = -5