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Question

The sum and product of three consecutive terms of an AP is given to be 21 and 168. If the common difference is known to be postitive, then the 10th term of the AP is ___.
  1. 52
  2. 38
  3. 45
  4. 58


Solution

The correct option is A 52
Let the terms of the AP be ad,a and a+d.
By question, we have
ad+a+a+d=21.
3a=21
a=7
Also, by question, we have (ad)a(a+d)=168.
i.e., (7d)7(7+d)=168
7(49d2)=168
3437d2=168
7d2=175
d2=25
d=±5
d>0d=5
The nth term of an AP of first term a and common difference d is given by
an=a+(n1)d.
a10=a+9d=7+9(5)=52
 

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