Find four numbers forming a GP. in which the third term is greater than the first by 9 and the second term is greater than the fourth by 18.

Solution

Let a, ar, ar², ar³ be the four numbers in G.P.
Since the third number is greater than the first by 9
∴ ar² = a + 9    $rightwards double arrow$ a(r² - 1) = 9 ...(i)
Since the second number is greater than the fourth by 18
∴ ar = ar³ + 18 $rightwards double arrow$ ar(r² - 1) = -18 ...(ii)
Dividing (ii) by (i), we get
   $fraction numerator ar left parenthesis straight r squared minus 1 right parenthesis over denominator straight a left parenthesis straight r squared minus 1 right parenthesis end fraction space equals space fraction numerator negative 18 over denominator 9 end fraction space space space space space space space space space space space space rightwards double arrow space space straight r space equals space minus 2$
Using r = -2 in (i), we get
   $straight a left square bracket left parenthesis negative 2 right parenthesis squared minus 1 right square bracket space equals space 9$ $rightwards double arrow$ a(4 - 1) = 9 $rightwards double arrow$ a = 3
∴ Numbers are 3, 3(-2), 3(-2)², 3(-2)³ or 3, -6, 12, -24

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Permutations And Combinations

Find four numbers forming a GP. in which the third term is greater than the first by 9 and the second term is greater than the fourth by 18.

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