Geometrical progression need help!?

If Tn = 2^(n + 3), then Tn-1 = 2^(n-1 + 3) = 2^(n + 2) so if we take the ratio Tn/Tn-1 we get

2^(n+3)/2^(n+2) and since a^m/a^n = a^(m-n), this becomes 2^1 = 2 i.e. the ratio of successive terms is a constant of 2) so the series is a G>P> {a "constant" is a fixed number (2 in this case) as opposed to a variable like x or n or whatever} to find the sum of the required we have to use the formula Sn = a(r^n - 1)/(r-1) where a = 1st term, r = common ratio (2 here) and n = number of terms.

Note that the formula only works for thw sum from 1 to n, so we have to use it twice: Sum to 12 terms - Sum to 5 terms gives us the sum from the 6th term to to 12th term.

First we need to find a. Tn = 2^(n+3) , so putting n = 1 we get a = 2^4 = 16 That gives us

S5 = 16 x (2^5 - 1)/(2-1) = 16 X 31 = 486

S12 = 16 x (2^12 -1)/(2 - 1) = 16 x 4095 = 16 x 4095 = 65,520 so the sum from the 6th to the 12th is

65,520 - 486 = 65,034

A constant is a number so that there is no n or any other variable invovled.

Tn=2^(n+3) and replacing n by (n-1) gives Tn-1 = 2^(n+2).

Tn/Tn-1 = 2^(n+3)/2^(n+2)=2^1=2, which is a constant since 2^p/2^q=26(p-q).

So the progression is geometric with common ratio r=2.

Also the first term T1 =a=2^4=16.

b) Find S5 and S12 using Sn=a(r^n - 1)/(r-1) = 16(2^n-1)

The required sum is S12 - S5 which should give 65024

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