In an A.P., the product of the first term and the second term is 120 and the product of the second term and the third term is 168. Find the tenth term of the A.P. when common difference d > 0.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 28
Writing the three terms around the middle term \(b\) as \(b-d\), \(b\), \(b+d\) makes the two products easy to combine.
\(b(b-d)=120\), that is, \(b^2-bd=120\) (1)
\(b(b+d)=168\), that is, \(b^2+bd=168\) (2)
Adding (1) and (2), the \(bd\) terms cancel: \(2b^2=288\), so \(b^2=144\)
Subtracting (1) from (2): \(2bd=48\), so \(bd=24\); with \(d\gt 0\), \(b\) is positive, so \(b=12\) and \(d=2\)
First term \(a=b-d=10\)
Tenth term \(=a+9d=10+18=28\)