Three positive intiger are in A.P …
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Rashmi Bajpayee 7 years, 2 months ago
Given, a1 + a2 + a3 = 33 ..........(i)
and a1 x a2 x a3 = 1155 ..........(ii)
Since a1, a2, a3 are in AP, therefore, a1 + a3 = 2a2 ..........(iii)
Substituting the value of a1 + a3 in eq. (i), we get 2a2 + a2= 33
=> a2= 11
Putting the value of a2 in eq. (i), we get,
a1 + a3 = 22
=> a1 = (22 - a3) .........(iv)
Putting the value of a2 in eq. (ii), we get, a1 x a3 = 105 .........(v)
Substituting the value of a1 in eq. (v), we get, (22 - a3) x a3 = 105
=> a32 - 22a3 + 105 = 0
=> a32 - 15a3 - 7a3 +105 = 0
=> a3(a3 - 15) - 7(a3 - 15) = 0
=> a3 - 15 = 0 and a3 - 7 = 0
=> a3 = 15 and a3 = 7
Putting these values in eq.(ii) we get, a1 = 7 and a1 = 15
Therefore, the required three positive integers are 7, 11 and 15.
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