Three numbers are in A.P. whose sum is 33 and product is 792 , then smallest number from these numbe

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8. Sequences and Series

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Three numbers are in $A.P.$ whose sum is $33$ and product is $792$, then the smallest number from these numbers is

A

$4$

B

$8$

C

$11$

D

$14$

Solution

(a) Suppose that three numbers are $a + d,\;a,\;a – d,$

therefore $a + d + a + a – d = 33$

$ \Rightarrow $$a = 11$

$a(a + d)(a – d) = 792$

$ \Rightarrow $$11(121 – {d^2}) = 792$

$ \Rightarrow $$d = 7$

Then required numbers are $4, 11, 18.$

Hence smallest number is $4.$

Standard 11

Mathematics

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