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 ___.52384558

Solution

## The correct option is A 52Let the terms of the AP be a−d,a and a+d. By question, we have a−d+a+a+d=21. ⇒3a=21 ⇒a=7 Also, by question, we have (a−d)a(a+d)=168. i.e., (7−d)7(7+d)=168 ⇒7(49−d2)=168 ⇒343−7d2=168 ⇒7d2=175 ⇒d2=25 ⇒d=±5 d>0⇒d=5 The nth term of an AP of first term a and common difference d is given by an=a+(n−1)d. ⇒a10=a+9d=7+9(5)=52

Suggest corrections