Question

3. Find four consecutive terms in an A.P. such that the sum of the middle two terms is 18 and product of the two end terms is 45. (Assume the four consecutive terms in A.P. are a-d,a,a+d,a+2d.)​

Answer 1

Step-by-step explanation:

Let the terms be a−d,a,a+d,a+2d

Sum=4a+2d=12

2a+d=6------------(1)

Sum of 3rd & 4th term=(a+d)+(a+2d)=14

2a+3d=14-----------(2)

By substracting Equation (1) from (2)

−2d=−8

⇒d=4

From Equation(1), a=1

So, the terms are (1−4),1,(1+4),(1+2(4))

=−3,1,5,9