Question

The first term and the 3rd term of a geometric sequence are 2 and 162 respectively.
(a) Find the general term Tn of the sequesnce.
(b) If the common ratio is less than 0, how many +ve terms are less than 15000?

Answer 1

@hartnn

Answer 2

\(\Large T_n = a_1 r^{n-1}\)

so, first term = a1 = 2

3rd term,
n=3
T3 = 162
\(\Large 162 = 2 r^{3-1}\)

firsr find 'r' from here.

Answer 3

r=9

Answer 4

T(2)=18 or -18?

Answer 5

we are done with part a, right ?
r= 9 or -9
Tn = 2 *(9)^{n-1}
or Tn = 2* (-9)^{n-1}

Answer 6

sorry, i would revise my statement for part b,

take r = -9.
(so alternate terms will be positive)

so, we will find total number of terms less than 15000
and divide it by 2 to get number of positive terms,
because half of the terms will be positive, and other half will be negative.

Answer 7

using calculator, n=6. But i don't have the correct answer.......

Answer 8

ok, we can do this, lets consider r as positive = +9,
and find the number of terms less than 15000

when r = -9, the only difference will be that half of those terms will be negative, so we can just divide that 'n' by 2 to get total number of positive terms when r is -9

does this make sense to you ??

2 (9)^{n-1} <15000
get n

Answer 9

but it says "if the common ratio is less than 0"

Answer 10

if r=+9, n=5.06088