A geometric series has 27 as the 1st term and a common ratio of 4/3. Find the least number of terms the...

... series can have if it exceeds 550?

advanced higher maths, please help? Show FULL working.


Thanks :)

Hi,

550 < 27(1 - (4/3)^x)/(1 - 4/3)

550 < 27(1 - (4/3)^x)/(-1/3)

550 < -81(1 - (4/3)^x)

-550/81 > 1 - (4/3)^x

-631/81 > -(4/3)^x

631/81 < (4/3)^x

log(631/81) < log(4/3)^x

log(631/81) < x log(4/3)

x > log(631/81)/log(4/3)

x> 7.13

If there are 8 or more terms, the sum is greater than 550. <==ANSWER

The sums of "n" terms are 27, 63, 111, 175, 260.3, 374.1, 525.8, and then 728.1.

So the 8th sum is greater than 550.

I hope that helps!! :-)

The sum of the first "n" terms of the series = 27(1 - (4/3)^n)/(1 - 4/3)

We need to find "n" where:
27(1 - (4/3)^n)/(1 - 4/3) > 550

27(1 - (4/3)^n)/(-1/3) > 550
-81(1 - (4/3)^n) > 550
81((4/3)^n - 1) > 550
(4/3)^n - 1 > 550/81
(4/3)^n > (550 + 81)/81
(4/3)^n > 631/81

Crunching the numbers with WolframAlpha: http://www.wolframalpha.com/input/?i=%28…

n > 7.13585…
n = 8

–––––––––––––––––––

Here are the first 8 terms:
27, 36, 48, 64, 256/3, 1024/9, 4096/27, 16384/81

Sum of the first 7 terms = 14197/27 = 525.815 approx
Sum of the first 8 terms = 58975/81 = 728.086 approx

–––––––––––––––––––

PI is correct if yu meant the SUM

I took it as the individual term. If so then

27*(4/3)^x > 550

(4/3)^x > 550/27

log(base 4/3)of(4/3)^x > log(base 4/3)of(550/27)

x > log(base 4/3)of(550/27)....................now use change of base formula

........log(550/27)
x > ------------------
........log(base 4/3)

x > 10.48...................MUST ROUND UP to next integer

x = 11.......ANSWER......11th term

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