if the first 3 terms of an AP are 3y-1, 3y+5, and 5y+1 respectively, find y
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a1= 3y-1, a2 =3y+5, a3= 5y+1.
Therefore, d or the common difference should be equal in each case.
d = a2/a1 = (3y+5)/3y-1 -----i)
Again, d= a3/a2 = (5y+1)/3y+5-----ii)
From equations i) and ii),
(3y+5)/3y-1 = (5y+1)/3y+5
=> (3y+5)² = (5y+1)(3y-1)
=> 9y²+30y+25 = 15y²-5y+3y-1
=> 9y²+30y+25 = 15y²-2y-1
=> 6y²-32y-26 = 0
=> 3y²-16y-13 = 0
Solve the equation using Sreedhar Acharya's formula, where a=3, b= -16, and c= -13 in the equation,
x = {-b ± √(b²-4ac)}/2a.
Therefore, d or the common difference should be equal in each case.
d = a2/a1 = (3y+5)/3y-1 -----i)
Again, d= a3/a2 = (5y+1)/3y+5-----ii)
From equations i) and ii),
(3y+5)/3y-1 = (5y+1)/3y+5
=> (3y+5)² = (5y+1)(3y-1)
=> 9y²+30y+25 = 15y²-5y+3y-1
=> 9y²+30y+25 = 15y²-2y-1
=> 6y²-32y-26 = 0
=> 3y²-16y-13 = 0
Solve the equation using Sreedhar Acharya's formula, where a=3, b= -16, and c= -13 in the equation,
x = {-b ± √(b²-4ac)}/2a.
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