In an arithmetic progression, the 25th term is 2552 and the 52nd term is 5279. What is the 79th term?
2007-05-23 08:45:23 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
If a(n) = c+nd then a(52) - a(25) = 52d - 25d = 27d
So d = (5279-2552)/27 = 101
a(79) = a(52) + (79-52)d = 5279 + 27*101 = 8006
2007-05-23 08:49:36 · answer #1 · answered by Astral Walker 7 · 0⤊ 0⤋
Arithmetic progression means you add a value to the number for every nth term. This means that between 25 and 52, the value was added 27 times (52-25). So:
5279 - 2552 = 27x
2727 = 27x
101 = x
So between 52 and 79, the value is added 27 times again.
x - 5279 = 27(101)
x - 5279 = 2727
x = 8006
2007-05-23 15:50:54 · answer #2 · answered by pjd4gnr 2 · 0⤊ 0⤋
lay it out
25, 2552, 52, 5279, Do you see anything that looks familiar?
The first number of the larger number is the number before. If you take 52-25 you get 27 and if you take 79-52 you get 27 so the number will be
79, 79106,
and so on
2007-05-23 15:50:23 · answer #3 · answered by tjnw79 4 · 0⤊ 0⤋
52 - 25 = 27
5279 - 2552 = 2727
interval = 2727/27 = 101
79 - 52 = 27
oh, for pete's sake that saves you a step.
In 27 intervals it increases 2727, so
5279 + 2727 = 8006
2007-05-23 15:49:00 · answer #4 · answered by TychaBrahe 7 · 0⤊ 1⤋
8006
2007-05-23 15:52:53 · answer #5 · answered by james g 1 · 0⤊ 0⤋
79106
2007-05-23 15:47:44 · answer #6 · answered by Anonymous · 0⤊ 0⤋