Since, a,b,181 are in GP, so
18a=b2
⇒a=18b2....(1)
Also, a1,10,b1 are in A.P., so
a1+b1=20
⇒a+b=20ab
⇒18b2+b=360b3
⇒360b2−18b−1=0[∵b=0]
⇒b=72018±324+1440
⇒b=72018+1764∵b>0
⇒b=121
⇒a=18×1441=81
Now, 16a+12b=16×81+12×121=3
For the two positive numbers a,b, if a,b and 181 are in a geometric progression, while a1,10 and b1 are in an arithmetic progression, then, 16a+12b is equal to _____ .
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