{"id":4076,"date":"2021-08-15T22:02:36","date_gmt":"2021-08-15T22:02:36","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4076"},"modified":"2021-11-17T17:45:56","modified_gmt":"2021-11-17T12:15:56","slug":"what-is-agp","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-agp\/","title":{"rendered":"What is AGP – How to Solve AGP Series"},"content":{"rendered":"
Here you will learn what is agp (Arithmetico Geometric Series) and how to solve agp series.<\/p>\n
Let’s begin –<\/p>\n
A series, each term of which is formed by multiplying the corresponding term of an A.P. & G.P. is called the Arithmetico-Geometric Series, e.g. 1 + 3x + 5\\(x^2\\) + 7\\(x^3\\) + ……..<\/p>\n
Here 1, 3, 5, ……. are in A.P. & 1, x, \\(x^2\\), \\(x^3\\) …….. are in G.P.<\/p>\n
Let \\(S_n\\) = a + (a + d)r + …….. + [a + (n-1)d]\\(r^{n-1}\\)<\/p>\n
then \\(S_n\\) = \\(a\\over {1-r}\\) + \\(dr({1}-{r}^{n-1})\\over (1-r)^2\\) – \\([a + (n-1)d]r^2\\over {1-r}\\), r \\(\\ne\\) 1<\/p>\n
If 0 < |r| < 1 & n \\(\\rightarrow\\) \\(\\infty\\)\u00a0 \u00a0\\(\\displaystyle \\lim_{x \\to \\infty}\\) \\(r^n\\) = 0, \\(S_{\\infty}\\) = \\(a\\over {1-r}\\) + \\(dr\\over (1-r)^2\\)<\/p>\n\n\n
Example : <\/span>Find the sum of series 4 – 9x + 16\\(x^2\\) – 25\\(x^3\\) + 36\\(x^4\\) – 49\\(x^5\\) + \n\t\t ……. \\(\\infty\\)<\/p>\n Solution : <\/span>Let S = 4 – 9x + 16\\(x^2\\) – 25\\(x^3\\) + 36\\(x^4\\) – 49\\(x^5\\) + ……. \\(\\infty\\) (a) \\({\\sum}_{r=1}^{n\u200e}\\)r = \\(n(n + 1)\\over 2\\) (sum of the first n natural numbers)<\/p>\n (b) \\({\\sum}_{r=1}^{n\u200e} r^2\\) = \\(n(n + 1)(2n + 1)\\over 6\\) (sum of the squares of the first n natural numbers)<\/p>\n (c) \\({\\sum}_{r=1}^{n\u200e} r^3\\) = \\(n^2(n + 1)^2\\over 4\\) (sum of the cubes of the first n natural numbers)<\/p>\n (d) \\({\\sum}_{r=1}^{n\u200e} r^4\\) = \\(n(n + 1)(2n + 1)(3n^2 + 3n -1)\\over 30\\)<\/p>\n (e) \\({\\sum}_{r=1}^{n\u200e}\\)(2r – 1) = \\(n^2\\) (sum of the first n odd natural numbers)<\/p>\n (f) \\({\\sum}_{r=1}^{n\u200e}\\)2r = n(n + 1) (sum of the first n even natural numbers)<\/p>\n Note :<\/strong><\/p>\n If \\(n^{th}\\) terms of a sequence is given by \\(T_n\\) = \\(an^3\\) + \\(bn^2\\) + cn + d where a, b, c, d are constants,<\/p>\n then sum of n terms \\(S_n\\) = \\(\\sum T_n\\) = a\\(\\sum n^3\\) + b\\(\\sum n^2\\) + c\\(\\sum n\\) + \\(\\sum d\\)<\/p>\n This can be evaluated using the above results.<\/p>\n\n\n Example : <\/span> Sum upto 16 terms of the series \\(1^3\\over 1\\) + \\(1^3 + 2^3\\over 1 + 3\\) + \n\t\t \\(1^3 + 2^3 + 3^3\\over 1 + 3 + 5\\) +…….. is<\/p>\n Solution : <\/span>\\(t_n\\) = \\(1^3 + 2^3 + 3^3 + …. + n^3\\over 1 + 3 + 5 + ….. + (2n – 1)\\)
\n\t\t -Sx = -4x + 9\\(x^2\\) – 16\\(x^3\\) + 25\\(x^4\\) – 36\\(x^5\\) + ……. \\(\\infty\\)
\n\t\t On Subtraction, we get
\n\t\t S(1 + x) = 4 – 5x + 7\\(x^2\\) – 9\\(x^3\\) + 11\\(x^4\\) – 13\\(x^5\\) + ……. \\(\\infty\\)
\n\t\t -S(1 + x)x = -4x + 5\\(x^2\\) – 7\\(x^3\\) + 9\\(x^4\\) – 11\\(x^5\\) + ……. \\(\\infty\\)
\n\t\t On Subtraction, we get
\n\t\t S\\((1 + x)^2\\) = 4 – x + 2\\(x^2\\) – 2\\(x^3\\) + 2\\(x^4\\) – 2\\(x^5\\) + ……. \\(\\infty\\)
\n\t\t = 4 – x + 2\\(x^2\\)(1 – x + \\(x^2\\) – \\(x^3\\) + ……. \\(\\infty\\)) = 4 – x + \\(2x^2\\over {1+x}\\)\n\t\t = \\({4 + 3x + x^2}\\over {1+x}\\)
\n\t\t S = \\({4 + 3x + x^2}\\over (1+x)^3\\)
\n<\/p>\n\n\nRESULTS<\/h4>\n
\n\t\t = \\({n^2(n + 1)^2\\over 4}\\over n\/2{[2+2(n – 1)]}\\) = \\({n^2(n + 1)^2\\over 4}\\over n^2\\)\n\t\t = \\({(n + 1)^2\\over 4}\\) = \\(n^2\\over 4\\) + \\(n\\over 4\\) + \\(1\\over 4\\)
\n\t\t \\(\\therefore\\) \\(S_n\\) = \\(\\sum t_n\\) + \\(1\\over 4\\)\\(\\sum n^2\\) + \\(1\\over 2\\)\\(\\sum n\\) + \\(1\\over 4\\)\\(\\sum 1\\)\n\t\t = \\(1\\over 4\\).\\(n(n + 1)(2n + 1)\\over 6\\) + \\(1\\over 2\\).\\(n(n + 1)\\over 2\\) + \\(1\\over 4\\).n
\n\t\t \\(\\therefore\\) \\(S_{16}\\) = \\(16.17.33\\over 24\\) + \\(16.17\\over 4\\) + \\(16\\over 4\\) = 446.
\n<\/p>\n\n\n\n