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 March 8th, 2016, 06:50 AM #1 Newbie   Joined: Mar 2016 From: Nigeria Posts: 1 Thanks: 0 Hi need help here: Another one please: The 1st, 2nd and 3rd terms of a GP are 1, 3 and 7 term of an AP respectively; if the second term of the AP is 30, find the sum of the first four terms of the GP. Last edited by skipjack; March 8th, 2016 at 06:56 AM. March 8th, 2016, 08:40 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,937 Thanks: 2210 Your first question was the same. Posting it once suffices. Let a, ar and ar² (in that order) be the first three terms of the GP. Any AP's 2nd term is the mean of its 1st and 3rd terms, so a + ar = 2 × 30 = 60. Hence a²r² = (60 - a)² = a² - 120a + 3600. Any AP's 7th term is 6 times its 2nd term minus 5 times its 1st term, so ar² = 6 × 30 - 5a. Hence a²r² = 180a - 5a². It follows that a² - 120a + 3600 = 180a - 5a², which implies 6a² - 300a + 3600 = 0, i.e. 6(a - 20)(a - 30) = 0. Hence a = 20 (as a = 30 would imply r = 1, which isn't allowed for a GP), and so r = 2. The required sum is therefore 20 + 40 + 80 + 160 = 300. Tags sequence, series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post naman Algebra 3 March 6th, 2013 08:22 PM Maths4ever Algebra 4 October 4th, 2012 01:35 AM suomik1988 Algebra 5 December 8th, 2010 07:38 AM danield3 Algebra 1 May 5th, 2010 07:15 PM suomik1988 Calculus 5 December 31st, 1969 04:00 PM

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