{"id":2891,"date":"2023-10-03T15:59:22","date_gmt":"2023-10-03T15:59:22","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2891"},"modified":"2023-10-03T16:03:34","modified_gmt":"2023-10-03T16:03:34","slug":"week-4-ss2-first-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-4-ss2-first-term-mathematics-notes\/","title":{"rendered":"Week 4 &#8211; SS2 First Term Mathematics Notes"},"content":{"rendered":"<p>\u00a0<strong>WEEK FOUR<br \/>\n<\/strong><strong>TOPIC:  GEOMETRIC PROGRESSION<br \/>\n<\/strong><strong>CONTENT<br \/>\n<\/strong><\/p>\n<ul>\n<li>Definition of Geometric Progression\n<\/li>\n<li>Denotations of Geometric progression\n<\/li>\n<li>The nth term of a G. P.\n<\/li>\n<li>The sum of Geometric series\n<\/li>\n<li>Sum of G. P. to infinity\n<\/li>\n<li>\n<div>Geometric mean\n<\/div>\n<p><strong>Definition of G. P<br \/>\n<\/strong>The sequence 5, 10, 20, 40 has a first term of 5 and the common ratio<br \/>\nBetween the term is 2 e.g. (<sup>10<\/sup>\/<sub>5<\/sub> or <sup>40<\/sup>\/<sub>2o<\/sub> = 2).\n<\/li>\n<\/ul>\n<p>A sequence in which the terms either increase or decrease in a common ratio is called a Geometric Progression<br \/>\n(G. P)<br \/>\nG. P:     a,      ar,     ar<sup>2<\/sup>,        ar<sup>3<\/sup> \u2026\u2026\u2026\u2026\u2026\u2026<\/p>\n<p>\u00a0Denotations in G. P<br \/>\na    = 1<sup>st<\/sup> term<br \/>\nr    = common ratio<br \/>\nU<sub>n<\/sub> = nth term<br \/>\nS<sub>n<\/sub>  = sum<\/p>\n<p>\u00a0<strong>The nth term of a G. P<br \/>\n<\/strong>The nth term = Un<br \/>\nUn = ar<sup>n-1<\/sup><br \/>\n\t\t1<sup>st<\/sup> term   = a<br \/>\n2<sup>nd<\/sup> term = a x r =ar<br \/>\n3<sup>rd<\/sup> term = a x r x r = ar<sup>2<\/sup><br \/>\n\t\t4<sup>th<\/sup> term = a x r x r x r =  ar<sup>3<\/sup><br \/>\n\t\t8<sup>th<\/sup>term  =  a x r x r x r x r x r x r x r  =  ar<sup>7<\/sup><br \/>\n\t\tnth term  =  a x r x r x r x \u2026\u2026\u2026.. <strong>ar<sup>n-1<\/sup><\/strong><\/p>\n<p>\u00a0<strong>Example<br \/>\n<\/strong>Given the GP 5, 10, 20, 40. Find its (a) 9<sup>th<\/sup> term (b) nth term<br \/>\nSolution<br \/>\na = 5         r  = 10\/5  =  2<br \/>\nU<sub>9<\/sub>\u00a0\u00a0\u00a0\u00a0=   ar<sup>n-1<\/sup><br \/>\n\t\tU<sub>9<\/sub>\u00a0\u00a0\u00a0\u00a0=   5 (2) <sup>9-1<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0=   5 (2)<sup>8<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0=   5 x 256   = 1,280<br \/>\n(b)\u00a0\u00a0\u00a0\u00a0U<sub>n<\/sub> = ar<sup>n-1<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0      = 5(2) <sup>n-1<\/sup><br \/>\n\t\t<strong>Example 2<br \/>\n<\/strong>The 8<sup>th<\/sup> term of a G.P is -7\/32. Find its common ratio if it first term is 28.<br \/>\nU<sub>8<\/sub>  =  -7\/32           U<sub>n<\/sub>        =  ar<sup>n-1<\/sup><br \/>\n\t\t -7\/32   =  28 (r)<sup>8-1<\/sup><br \/>\n\t\t -7\/32   =  28r<sup>7<\/sup><br \/>\n\t\t -7\/32 x 1\/28 =<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi1.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi2.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi3.png\" alt=\"\"\/>  &#8211;<sup>7<\/sup>\/<sub>32<\/sub> x <sup>1<\/sup>\/<sub>28<\/sub> = r<sup>7<br \/>\n<\/sup><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi4.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi5.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi6.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi7.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi8.png\" alt=\"\"\/><sup><br \/>\n\t\t\t<\/sup><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi9.png\" alt=\"\"\/>           7                                       7<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi10.png\" alt=\"\"\/>r =                              =          <\/p>\n<p>\u00a0   r = &#8211; 0.5<\/p>\n<p>\u00a0<strong>Evaluation<br \/>\n<\/strong>1.    The 6<sup>th<\/sup> term of a G.P is 2000. Find its first term if its common ratio is 10.<br \/>\n2.    Find the  7<sup>th<\/sup>  term   and  the   nth   term   of  the   progression   27, 9 , 3, \u2026 <\/p>\n<p>\u00a0<strong>THE SUM OF A GEOMETRIC SERIES<br \/>\n<\/strong>a + ar + ar<sup>2<\/sup> + ar<sup>3<\/sup> + \u2026\u2026\u2026\u2026\u2026\u2026. ar<sup>n-1<\/sup><br \/>\n\t\trepresent a general geometric series where the terms are added.<br \/>\nS =  a + ar + ar<sup>2<\/sup> \u2026\u2026\u2026\u2026 ar<sup>n-1 <\/sup>eqn 1<br \/>\nMultiply through r<br \/>\nrs =  ar + ar<sup>2<\/sup> + ar<sup>3<\/sup> \u2026\u2026\u2026. ar<sup>n<\/sup> \u2026\u2026\u2026 eqn 2<br \/>\nsubtracteqn 2 from 1<\/p>\n<p>\u00a0<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi11.png\" alt=\"\"\/>S \u2013 rs = a \u2013 ar<sup>n<\/sup><br \/>\n\t\tS (1 \u2013 r)  =a(1-r<sup>n<\/sup>)<br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi12.png\" alt=\"\"\/>    1 \u2013 r           1-r<\/p>\n<p>\u00a0<strong>S.=a ( 1 &#8211; r<sup>n<\/sup>)<\/strong>         r &lt; 1<br \/>\n<strong>1 &#8211; r<br \/>\n<\/strong>Multiply through by -1 or subs. eqn. 1 from e.g. 2<br \/>\nrs  &#8211;  s  =  ar<sup>n<\/sup> \u2013 a<br \/>\nS (r \u2013 1)  =a(r<sup>n<\/sup> \u2013 1)<br \/>\n\t\t    r \u2013 1\u00a0\u00a0\u00a0\u00a0          r \u2013 1<br \/>\n   S =     <strong>a(r<sup>n<\/sup>-1)<\/strong><br \/>\n\t\t<strong>r -1<\/strong>            for r &gt; 1<\/p>\n<p>\u00a0<strong>Example:<br \/>\n<\/strong>Find the sum of the series.<br \/>\na.\u00a0\u00a0\u00a0\u00a0\u00bd + \u00bc + 1\/8 + \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 as far as 6<sup>th<\/sup> term<br \/>\nb.\u00a0\u00a0\u00a0\u00a01 + 3 + 9 + 27 + \u2026\u2026\u2026\u2026\u2026\u2026\u2026. 729<br \/>\n<strong>Solution<br \/>\n<\/strong>a  =  \u00bd<br \/>\nr  =  \u00bd      (r = \u00bc \u00b8 \u00bd = \u00bd)<br \/>\n\\r&lt; 1<br \/>\nS = a (1-r<sup>n<\/sup>)<br \/>\n\t\t        1 \u2013 r<br \/>\nS<sub>6<\/sub>  =  [\u00bd (1 \u2013 (\u00bd)<sup>6<\/sup>]<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0        1 &#8211; \u00bd<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi13.png\" alt=\"\"\/>S<sub>6<\/sub>=  \u00bd (1 \u2013 <sup>1<\/sup>\/<sub>64<\/sub>)<br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi14.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00bd<br \/>\nS<sub>6 <\/sub>  = 1 \u2013 1   = 64 \u2013 1   = 63<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0 64       64          64<br \/>\n2.\u00a0\u00a0\u00a0\u00a0a = 1, r = 3, n = ?  U<sub>n<\/sub> = 729<br \/>\n\u00a0\u00a0\u00a0\u00a0U<sub>n<\/sub> = ar<sup>n-1<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0729 = 1 x 3<sup>n-1<\/sup>             (3<sup>n-1<\/sup> = 3<sup>n<\/sup> x 3<sup>-1<\/sup>)<br \/>\n\u00a0\u00a0\u00a0\u00a0729 = 3n<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0           3<br \/>\n\u00a0\u00a0\u00a0\u00a03<sup>n<\/sup> = 3 x 729<br \/>\n\u00a0\u00a0\u00a0\u00a03<sup>n<\/sup> = 31 x 36<br \/>\n\u00a0\u00a0\u00a0\u00a03<sup>n<\/sup> = 3<sup>7<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\\ n = 7<br \/>\n\u00a0\u00a0\u00a0\u00a0S = a(r<sup>n<\/sup>-1)<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0        r \u2013 1<br \/>\n\u00a0\u00a0\u00a0\u00a0S = a(3<sup>7<\/sup> \u2013 1)    =   2187 &#8211; 1<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0        3 \u2013 1                  2<br \/>\n\u00a0\u00a0\u00a0\u00a02186   = 1093<br \/>\n                     2<\/p>\n<p>\u00a0<strong>Evaluation: <\/strong>Find the sum of the series 40, -4, 0.4 as far as the 7<sup>th<\/sup> term.<strong><br \/>\n\t\t\t<\/strong><br \/>\n\u00a0<strong>SUM OF G. P. TO INFINITY<br \/>\n<\/strong>Sum of G. P to infinity is only possible where r is &lt; 1.<br \/>\nWhere r is &gt; 1 there is no sum to infinity.<br \/>\nExample:<br \/>\n1. Find the sum of G. P. 1 + \u00bd + \u00bc + \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (a) to 10 terms    (b) to 100 terms. Hence deduce the sum of the series (formula) if it has a very large no. of term or infinity.<br \/>\n(a)\u00a0\u00a0\u00a0\u00a0a = 1   r = \u00bd<br \/>\n\u00a0\u00a0\u00a0\u00a0n = 10<br \/>\n\u00a0\u00a0\u00a0\u00a0S = a (1-r<sup>n<\/sup>)<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0          1-r<br \/>\n\u00a0\u00a0\u00a0\u00a0S = 1(1-(<sup>1<\/sup>\/<sub>2<\/sub>)<sup>10<\/sup>) =   1(1-0.0001)<br \/>\n                       1- \u00bd                    1\/2<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02 (1 \u2013 0.001)<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02 \u2013 0.002  =  1.998.<br \/>\nb.\u00a0\u00a0\u00a0\u00a0n = 100.<br \/>\n\u00a0\u00a0\u00a0\u00a0S = a (1 \u2013 r<sup>n<\/sup>)<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0          1 &#8211; r<br \/>\n\u00a0\u00a0\u00a0\u00a0S =  1 (1-(<sup>1<\/sup>\/<sub>2<\/sub>)<sup>100<\/sup>)  =   1(1- (<sup>1<\/sup>\/<sub>2<\/sub>)<sup>10<\/sup>)<sup>10<\/sup><br \/>\n\t\t\t\u00a0\u00a0\u00a0\u00a0         1 \u2013 \u00bd \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0     \u00bd<br \/>\n\u00a0\u00a0\u00a0\u00a01 (1-(0.001)<sup>10<\/sup><br \/>\n\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00bd<br \/>\n\u00a0\u00a0\u00a0\u00a01 (1)<br \/>\n\u00a0\u00a0\u00a0\u00a0    \u00bd  \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= <strong>2<\/strong><br \/>\n\t\tTherefore (1\/2)<sup>100<\/sup> tend to 0 (infinity).<\/p>\n<p>\u00a0In general,<br \/>\nS = a (1-r<sup>n<\/sup>)=  a(1-0)  =     a__<br \/>\n\u00a0\u00a0\u00a0\u00a01-r\u00a0\u00a0\u00a0\u00a01-r\u00a0\u00a0\u00a0\u00a01 \u2013 r<br \/>\n\\ S<sub>\u00a5<\/sub><strong>=    a__<\/strong>    =    n \u2192<br \/>\n\t\t\t\u00a5<br \/>\n\t\t\t\u00a0\u00a0\u00a0\u00a0<strong>1 \u2013 r<br \/>\n<\/strong><strong>Example 2:<br \/>\n<\/strong>Find the sum of the series 45 + 30 + 20 + \u2026\u2026\u2026\u2026\u2026\u2026 to infinity.<br \/>\na =  45,  r = <sup>2<\/sup>\/<sub>3<\/sub>,  n = infinity<br \/>\nS\u221e =    a                    S =    45__<br \/>\n       1 \u2013 r                          1- <sup>2<\/sup>\/<sub>3<\/sub><br \/>\n\t\tS\u221e  =  45 \u00f7 <sup>1<\/sup>\/<sub>3<\/sub><br \/>\n\t\t         45 x <sup>3<\/sup>\/<sub>1<\/sub><br \/>\n\t\t=  <strong>135<br \/>\n<\/strong><br \/>\n\u00a0<strong>Evaluation<br \/>\n<\/strong>1.\u00a0\u00a0\u00a0\u00a0The sum to infinity of a Geometric Series is 100. Find the first term if the common ratio is &#8211;<sup>1<\/sup>\/<sub>2<\/sub>.<br \/>\n2.\u00a0\u00a0\u00a0\u00a0The 3<sup>rd<\/sup> and 6<sup>th<\/sup> term of a G. P. are 48 and 14<sup>2<\/sup>\/<sub>9<\/sub> respectively, write down the first four terms of the G. P.<br \/>\n3.\u00a0\u00a0\u00a0\u00a0The sum of a G. P. is 100 find its first term if the common ratio is 0.8.<\/p>\n<p>\u00a0<strong>GEOMETRIC MEAN<br \/>\n<\/strong>If three numbers such as x , y and z are consecutive terms of a G.P then their common ratio will be<br \/>\ny  =z<br \/>\n\t\tx      y<br \/>\n                              y<sup>2<\/sup> = xz<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi15.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi16.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi17.png\" alt=\"\"\/>                              y =    xz<br \/>\nThe middle value , y is the geometric mean (GM). We can conclude by saying that the GM of two numbers is the positive square root of their products.<br \/>\n<strong>Example<br \/>\n<\/strong>Calculate the geometric mean of I. 3 and 27    II. 49 and 25<br \/>\n\t\t                                                                                           4<br \/>\n<strong>    Solution<br \/>\n<\/strong><\/p>\n<ol>\n<li><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi18.png\" alt=\"\"\/>G.M of 3 and 27                 II.  G.M of  49 and 25\n<\/li>\n<\/ol>\n<p>=  \u221a 3 x 27                                                    4<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi19.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi20.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100323_1559_Week4SS2Fi21.png\" alt=\"\"\/>                  =   \u221a 81                                               =     49 x 25<br \/>\n\t\t                  =    9                                                                    4<br \/>\n                                                                              =   7 x 5<br \/>\n\t\t                                                                                         2<br \/>\n                                                                               =   35     = 17 1\/2<br \/>\n                                                                                     2<br \/>\n<strong>Example<br \/>\n<\/strong>The first three terms of a GP are k + 1, 2k \u2013 1, 3k + 1. Find the possible values of the common ratio.<br \/>\nSolution<br \/>\nThe terms are  k + 1, 2k \u2013 1, 3k + 1<br \/>\n2k -1  =3k + 1<br \/>\n\t\t                                     k + 1       2k \u2013 1<br \/>\n                                   (2k-1)(2k-1) = (k+1)(3k+1)<br \/>\n                               4k<sup>2<\/sup>-2k-2k +1 = 3k<sup>2<\/sup> +k+3k + 1<br \/>\n                               4k<sup>2<\/sup>&#8211; 4k +1   = 3k<sup>2<\/sup> +4k + 1<br \/>\n                                4k<sup>2 <\/sup>&#8211; 3k<sup>2 <\/sup>&#8211; 4k &#8211; 4k + 1-1 = 0<br \/>\nk<sup>2<\/sup> -8k = 0<br \/>\nk(k-8) = 0<br \/>\n                                       k = 0 or k &#8211; 8 = 0<br \/>\n                                       k = 0 or 8<br \/>\nThe common ratio will have two values due to the two values of k<br \/>\nWhen k=0                                                       when k= 8<br \/>\nK+1 = 0+1 =1                                               k+1 = 8+1 = 9<br \/>\n2k- 1= 2&#215;0 \u2013 1 = -1                                               2k- 1 = 2&#215;8 \u2013 1 = 15<br \/>\n3k+ 1= 3&#215;0+ 1 = 1                                                3k+1 = 3&#215;8 +1 = 25<br \/>\nterms are 1 , -1 , 1                                                 terms are 9,15,25<br \/>\ncommon ratio, r = -1\/1                                      common ratio,r = 15\/9<br \/>\n                         r = -1<br \/>\n<strong>EVALUATION<br \/>\n<\/strong>The third term of a G.P. is 1\/81.  Determine the first term if the common ratio is 1\/3.<\/p>\n<p>\u00a0<strong>GENERAL EVALUATION \/REVISION QUESTION<br \/>\n<\/strong>1. p &#8211; 6, 2p  and 8p + 20 are three consecutive terms of a GP. Determine the value of   (a) p (b) the common ratio<br \/>\n2. If 1  , x , 1 , y , \u2026.are in GP , find the product of x and y<br \/>\n       16        4<br \/>\n3.The   third  term  of  a  G.P   is  45  and  the  fifth  term  405.Find  the  G.P. if  the  common  ratio  r  is  positive.<br \/>\n4.Find  the  7<sup>th<\/sup>  term  and  the  nth  term  of  the  progression  27,9,3,\u2026<br \/>\n5.In  a  G.P, the  second  and  fourth  terms  are  0.04  and  1 respectively. Find the (a) common ratio (b) first term<\/p>\n<p>\u00a0<strong>WEEKEND ASSIGNMENT<br \/>\n<\/strong>1. In the 2<sup>nd<\/sup> and 4<sup>th<\/sup> term of a G.P are 8 and 32 respectively, what is the sum of the first four terms.  (a) 28    (b)  40    (c)  48    (d) 60<br \/>\n2. The sum of the first five term of the G.P. 2, 6, 18, is     (a) 484    (b) 243      (c) 242    (d) 130<br \/>\n3. The 4<sup>th<\/sup> term of a GP is   -2\/3 and its first term is 18 what is its common ratio.  (a) \u00bd    (b) <sup>1<\/sup>\/<sub>3<\/sub><br \/>\n\t\t(c) <sup>-1<\/sup>\/<sub>3<\/sub>   (d)  <sup>-1<\/sup>\/<sub>2<\/sub><br \/>\n\t\t4. If the 2<sup>nd<\/sup> and 5<sup>th<\/sup> term of a G. P. are -6 and 48 respectively, find the sum of the first four terms:  (a) -45    (b) -15     (c) 15    (d) 33<br \/>\n5. Find the first term of the G.P. if its common ratio and sum to infinity \u2013 <sup>3<\/sup>\/<sub>3<\/sub> and respectively (a) 48    (b) 18    (c) 40   (d) -42 <\/p>\n<p>\u00a0<strong>THEORY<br \/>\n<\/strong>1.The 3<sup>rd<\/sup> term of a GP is 360 and the 6<sup>th<\/sup> term is 1215. Find the<strong><br \/>\n\t\t\t<\/strong>(i)\u00a0\u00a0\u00a0\u00a0Common ratio         (ii)\u00a0\u00a0\u00a0\u00a0First term        (iii)\u00a0\u00a0\u00a0\u00a0Sum of the first four terms<br \/>\n1b. If (3- x) + (6) + (7- 5x) is a geometric series, find two possible values for<br \/>\n(i)   x     (ii)  the  common  ratio,  r          (iii) the sum  of  the   G.P<br \/>\n2.The first term of a G. P. is 48. Find the common ratio between its terms if its sum to infinity is 36.<\/p>\n<p>\u00a0<strong>Reading Assignment<br \/>\n<\/strong>New General Mathematics SSS2 <\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<br \/>\n\u00a0<strong><br \/>\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00a0WEEK FOUR TOPIC: GEOMETRIC PROGRESSION CONTENT Definition of Geometric Progression Denotations of Geometric progression The&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,233],"tags":[],"class_list":["post-2891","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss2-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2891","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2891"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2891\/revisions"}],"predecessor-version":[{"id":2892,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2891\/revisions\/2892"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2891"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2891"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2891"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}