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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.

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