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March 8th, 2016, 06:50 AM   #1
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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.
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March 8th, 2016, 08:40 AM   #2
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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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