The sum of the first three terms of an AP is 42. If the product of the first and third term exceeds the second term by 157,then the 10th term of the AP can be- A)59. B)49 C)54. D)44
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Let the first three terms of AP is a−d,a,a+d.
According to problem,
a−d+a+a+d=33 →(1)
(a−d)(a+d)=a+29→(2)
From (1), we get
3a=33a=11
From (2), we get
a 2 −d 2 =a+29→(3)
Put a=11 in equation (3)
121−d 2 =11+29121−d 2 =40
121−40=d 281=d 2d=±9
When a=11 & d=9
A.P: 2,11,20,29,.....
When a=11 & d=−9
A.P: 20,11,2,......
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ap 20 11 2...
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