find four consecutive terms AP whose sum is 66 and the product of the extremes to the product of the means of 14 : 15 . (the terms are in ascending order)
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let the terms be (a-3d), (a-d), (a+d) and (a+3d)
the sum is 66
a-3d+a-d+a+d+a+3d=66
4a=66
a=33/2
product of extremes to product of means is 14:15
(a-3d)(a+3d)/(a-d)(a+d)=14/15
solve this by substituting a=33/2
the sum is 66
a-3d+a-d+a+d+a+3d=66
4a=66
a=33/2
product of extremes to product of means is 14:15
(a-3d)(a+3d)/(a-d)(a+d)=14/15
solve this by substituting a=33/2
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