Here you will learn what is geometric progression (gp) and sum of gp series formula and properties of gp.<\/p>\n
Let’s begin –<\/p>\n
G.P. is a sequence of non zero numbers each of the succeeding term is equal to the preceding term multiplied by a constant. Thus in GP the ratio of successive terms is constant. This constant factor is called the COMMON RATIO of the sequence & is obtained by dividing any term by the immediately previous term. Therefore a, ar, a\\(r^2\\), ……. is a GP with ‘a’ as the first term and ‘r’ as common ratio.<\/p>\n
\n\\(T_n\\) = a\\(r^{n-1}\\)<\/p>\n<\/blockquote>\n
(b)\u00a0 Sum of the first n terms of GP<\/h3>\n
\n\\(S_n\\) = \\(a(r^n – 1)\\over {r – 1}\\), if r \\(\\ne\\) 1<\/p>\n<\/blockquote>\n
(c)\u00a0 Sum of infinite G.P. or sum of gp to infinity<\/a>,\u00a0<\/h3>\n
\n\\(S_n\\) = \\(a\\over {1 – r}\\); 0 < |r| < 1\u00a0<\/p>\n<\/blockquote>\n\n\n
Example : <\/span>Find the sum of 7 terms of the gp 3,6, 12, ……<\/p>\n
Solution : <\/span>Here a = 3, r = 2
\n\t\t \\(S_7\\) = \\(a(r^7 – 1)\\over {r – 1}\\)
\n\t\t \\(T_n\\) = \\(3(2^7 – 1)\\over {2 – 1}\\)
\n\t\t = 3(128 – 1)
\n\t\t = 381
\n\t\t <\/p>\n\n\nProperties of GP<\/h2>\n
(a)\u00a0 If each term of a G.P. be multiplied or divided by the some non-zero quantity, then the resulting sequence is also a G.P.<\/p>\n
(b)\u00a0 Three consecutive terms of a G.P. : a\/r, a, ar;<\/p>\n
Four consecutive terms of a G.P. : \\(a\/r^3\\), a\/r, ar \\(ar^3\\)<\/p>\n
(c)\u00a0 If a, b, c are in G.P. then \\(b^2\\) = ac.<\/p>\n
(d)\u00a0 If in a G.P, the product of two terms which are equidistant from the first and the last term, is constant and is equal to the product of first and last term. => \\(T_k\\).\\(T_{n-k+1}\\) = constant = a.l<\/p>\n
(e)\u00a0 If each term of a G.P. be raised to the same power, then resulting sequence is also a G.P.<\/p>\n
(f)\u00a0 In a G.P., \\({T_r}^2\\) = \\(T_{r-k}\\).\\(T_{r+k}\\), k < r, r \\(\\ne\\) 1<\/p>\n
(g)\u00a0 If the terms of a given G.P. are chosen at regular intervals, then the new sequence is also a G.P.<\/p>\n
(h)\u00a0 If \\(a_1\\), \\(a_2\\), \\(a_3\\)……\\(a_n\\) is a G.P. of positive terms, then log \\(a_1\\), log \\(a_2\\)……log \\(a_n\\) is an A.P. and vice-versa.<\/p>\n
(i)\u00a0 If \\(a_1\\), \\(a_2\\), \\(a_3\\)…… and \\(b_1\\), \\(b_2\\), \\(b_3\\)…… are two G.P.’s then \\(a_1\\)\\(b_1\\), \\(a_2\\)\\(b_2\\), \\(a_3\\)\\(b_3\\)……. & \\(a_1 \\over b_1\\), \\(a_2 \\over b_2\\), \\(a_3 \\over b_3\\)……. is also in G.P.<\/p>\n\n\n
Example : <\/span> If a, b, c, d and p are distinct real numbers such that\n\t\t\t \\((a^2 + b^2 + c^2)p^2\\) – 2p(ab + bc + cd) + \\((b^2 + c^2 + d^2)\\) \\(\\leq\\) 0 then a, b, c, d are in<\/p>\n
Solution : <\/span>Here, the given condition \\((a^2 + b^2 + c^2)p^2\\) – 2p(ab + bc + cd) + \\((b^2 + c^2 + d^2)\\) \\(\\leq\\) 0
\n\t\t => \\((ap – b)^2\\) + \\((bp – c)^2\\) + \\((cp – d)^2\\) \\(\\leq\\) 0
\n\t\t \\(\\because\\) a square cannot be negative
\n\t\t \\(\\therefore\\) ap – b = 0, bp – c = 0, cp – d = 0
\n\t\t => p = \\(b \\over a\\) = \\(c \\over b\\) = \\(d \\over c\\) => a, b, c, d are in G.P.
<\/p>\n\n\n
\nRelated Questions<\/h3>\n
The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, \u2026\u2026. , is<\/a><\/p>\n