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If ‘a’ is the first term and ‘r’ is the common ratio of the geometric progression, then its nth term is given by an = arn-1
The sum, Sn of the first ‘n’ terms of the G.P. is given by Sn = a (rn – 1)/ (r-1), when r ≠1; = na , if r =1
If -1 < x < 1, then limxn = 0, as n →∞. Hence, the sum of an infinite G.P. is 1+x+x2+ ….. = 1/(1-x)
If -1 < r< 1, then the sum of the infinite G.P. is a +ar+ ar2+ ….. = a/(1-r)
If each term of the G.P is multiplied or divided by a non-zero fixed constant, the resulting sequence is again a G.P.
If a1, a2, a3, …. andb1, b2, b3, … are two geometric progressions, then a1b1, a2b2, a3b3, …… is also a geometric progression and a1/b1, a2/b2, ... ... ..., an/bn will also be in G.P.
Suppose a1, a2, a3, ……,an are in G.P. then an, an–1, an–2, ……, a3, a2, a1 will also be in G.P.
Taking the inverse of a G.P. also results a G.P. Suppose a1, a2, a3, ……,an are in G.P then 1/a1, 1/a2, 1/a3 ……, 1/an will also be in G.P
If we need to assume three numbers in G.P. then they should be assumed as a/b, a, ab (here common ratio is b)
Four numbers in G.P. should be assumed as a/b3, a/b, ab, ab3 (here common ratio is b2)
Five numbers in G.P. a/b2, a/b, a, ab, ab2 (here common ratio is b)
If a1, a2, a3,… ,an is a G.P (ai> 0 ∀i), then log a1, log a2, log a3, ……, log an is an A.P. In this case, the converse of the statement also holds good.
If three terms are in G.P., then the middle term is called the geometric mean (G.M.) between the two. So if a, b, c are in G.P., then b = √ac is the geometric mean of a and c.
Likewise, if a1, a2, ……,an are non-zero positive numbers, then their G.M.(G) is given by G = (a1a2a3 …… an)1/n.
If G1, G2, ……Gn are n geometric means between and a and b then a, G1, G2, ……,Gn b will be a G.P.
Here b = arn+1, ⇒ r = n+1√b/a, Hence, G1 = a. n+1√b/a, G2 = a(n+1√b/a)2,…, Gn = a(n+1√b/a)n.
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Solved Examples on Geometric Progression...