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5! or 5P5 . A total of 120 series
I'll try to answer the question, "If the 5th term of a geometric progression is 2, then the product of its FIRST 9 terms is --?" Given the first term is A and the ratio is r, then the progression starts out... A, Ar, Ar^2, Ar^3, Ar^4, ... So the 5th term is Ar^4, which equals 2. The series continues... Ar^5, Ar^6, Ar^7, Ar^8, ... Ar^8 is the 9th term. The product P of all 9 terms is therefore: P = A * Ar * Ar^2 *...*Ar^8 Collect all the A's P = (A^9)*(1 * r * r^2 ...* r^8) P = A^9 * r^(0+1+2+...+8) There's a formula for the sum of the first n integers (n/2)(n+1), or if you don't know just add it up. 1+2+...+8 = 36 Therefore P = A^9 * r^36 Since 36 is a multiple of 9, you can simplify: P = (Ar^4)^9 Still with me? Remember that Ar^4=2 (a given fact). So finally P = 2^9 = 512. Cute problem.
There is no answer as expressed, since it is not a question. Are you trying to simplify it? 9r2s + 4rs + 5s2 - 9s2 - 9r2 - 5rs - s2 + 2r2s + 4rs + 2s2 = (9 + 2)r2s + (4 - 5 + 4)rs + (5 - 9 - 1 + 2)s2 = 11r2s + 3rs -3s2 = s(11r2 + 3r -3s)
there ar about 200- 100 -_200
This is a geometric sequence of the form a, ar, ar^2, ar^3, ... where a is the first term and r is the common ratio.In our case, the first term a = 2, and the common ratio r = 5.The nth term of such a sequence isan = a r^(n -1).
(Kr)5s2 4d10 5p5
I
[Kr] 5s2 4d10 5p5
1s22s22p63s23p64s2 3d104p65s24d105p5
long hand: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5 short hand: [Kr] 5s2 4d10 5p5 *remember the number after the letter is written as an exponent*
[Kr] 4d10 5s2 5p5
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5
It is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5.
(Kr)5s2 4d10 5p5
The electronic configuration of iodine is: [Kr] 4d10 5s2 5p5.
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6