1.

In an A.P. the sum of three consecutive terms is 27 and their product is 504. Find the terms. ( Consider the terms to be in ascending order. )

Answer» Correct Answer - The three consecutive terms are 4,9 and 14.
Let the three consecutive terms in A.P. be a-d,a and a+d.
From the first condition.
a-d+a+a+d= 27
`:. 3a = 2` `:. A= 9` …..(1)
From the second condition.
`(a-d) (a) (a+d) = 504`
`:. a(a^(2) -d^(2))= 504`
`:. 9((9)^(2) - d^(2)) = 504 ` ...[From (1)]
`:. 81 -d^(2) = 56` ...(Dividing both the sides by 9)
`:. d^(2) = 81-56 ` `:. d^(2) = 25 ` `:. d +- 5` ......(Taking square root )
But the terms are in ascending order, `:. d = 5`
Substituting a= 9 and d=5,a = 9-5=4, a = 9 , `
`a+d = 9+5= 14`


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