Answer :

To find: Three numbers


Given: Three numbers are in A.P. Their sum is 21


Formula used: When a,b,c are in GP, b2 = ac


Let the numbers be a - d, a, a + d


According to first condition


a + d + a +a – d = 21


3a = 21


a = 7


Hence numbers are 7 - d, 7, 7 + d


When second number is reduced by 1 and third is increased by 1 then the numbers become –


7 – d , 7 – 1 , 7 + d + 1


7 – d , 6 , 8 + d


The above numbers are in GP


Therefore, 62 = (7 – d) (8 + d)


36 = 56 + 7d – 8d – d2


d2 + d – 20 = 0


d2 + 5d – 4d – 20 = 0


d (d + 5) – 4 (d + 5) = 0


(d – 4) (d + 5) = 0


d = 4, Or d = -5


Taking d = 4, the numbers are


7 - d, 7, 7 + d = 7 - 4, 7, 7 + 4


= 3, 7, 11


Taking d = -5, the numbers are


7 - d, 7, 7 + d = 7 – (-5), 7, 7 + (-5)


= 12, 7, 2


Ans) We have two sets of triplet as 3, 7, 11 and 12, 7, 2.


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