Question:
If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
Answer:
The three numbers in A.P. is taken as a-d, a, a+d where a is the first term and d is the common difference of the series
Now sum of three numbers = 24
=> (a-d)+a+(a+d)= 24
=>3a = 24
=>a = 24/3
=>a = 8
Again product of three numbers = 440
=>(a-d)*a*(a+d) = 440
=>a*(a2 -d2 ) = 440
=>8*(64 - d2 ) = 440
=>64 -d2 =440/8
=>64 - d2 = 55
=>d2 = 64-55
=> d2 = 9
=>d = √9
=>d = 3, -3
1. When a=8, d=3 then the three numbers are
a-d, a, a+d = 8-3, 8, 8+3
= 5,8,11
1. When a=8, d=-3 then the three numbers are
a-d, a, a+d = 8-(-3), 8, 8+(-3)
= 8+3, 8, 8-3
= 11, 8, 5
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