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Grade 12Algebra

four no.s are in a.p. the sum of the 1st and last terms is 18 and the product of both the middle terms is 15 . the least no. of the series is.

Profile image of suvarna
8 Years agoGrade 12
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1 Answer

Profile image of Samyak Jain
8 Years ago
Let the 4 numbers in A.P. be a – 3d, a – d, a + d, a + 3d.
Given : (a – 3d) + (a + 3d) = 18  &  (a – d)(a + d) = 15
\therefore 2a = 18  =>  a = 9   &
a2 – d2 = 15  => 92 – d2 = 15  => d2 = 81 – 15 = 66
\therefore d = \pm\sqrt{}66
When d = \sqrt{}66 ,
a – 3d = 9 – 3\sqrt{}66  ,     a – d = 9 – \sqrt{}66
a + d = 9 + \sqrt{}66   ,        a + 3d = 9 + 3\sqrt{}66.
When d = – \sqrt{}66 , 
a – 3d = 9 + 3\sqrt{}66  ,     a – d = 9 + \sqrt{}66
a + d = 9 – \sqrt{}66   ,        a + 3d = 9 – 3\sqrt{}66.
\therefore The AP is either
9 – 3\sqrt{}66 , 9 – \sqrt{}66 , 9 + \sqrt{}66 , 9 + 3\sqrt{}66  or
9 + 3\sqrt{}66 , 9 + \sqrt{}66 , 9 – \sqrt{}66 , 9 – 3\sqrt{}66 .
Thus, the least number of the AP is 9 – 3\sqrt{}66.

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