sum of 3term of an ap is 48then the middle term is
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Let the first three terms of the series be.
a−d,a,a+d
Where d is the common difference.
Now as per the problem.
[math]a-d + a + a+d =48
3a=48
a=16[/math]
Product of the first and second term exceeds the 4 times the third term by 12 ie
[math](a-d)*a=4*(a+d)+12
Substitute a=16 in above equation
(16-d)*16=4*(16+d) + 12
256–16d=76 +4d
20d=256–76
d=180/20
d=9[/math]
Thus our required series is
16–9,16,16+9
Which is 7,16,25
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