Find four consecutive terms in an A.P. whose sum is 36 and the product

1.	Find four consecutive terms in an A.P. whose sum is 36 and the product of the 2nd and the 4th is 105. The terms are in ascending order.
Answer» ong>ANSWER: Step-by-step explanation: Let the FOUR consecutive TERMS of AP are a-3d,a-d,a+d and a+3d Sum = a-3d+a-d+a+d +a+3d=36 4a=36 a=9 Now the product of the 2ND and the 4th term is 105 So (a-d)(a+3d)=105 (9-d)(9+3d)=105 $81+27d-9d-3d^2=105$ $3d^2-18d-81=105\\\\d^2-6d-+8=0\\\\d^2-4d-2d+8=0\\\\d(d-4)-2(d-4)=0\\\\(d-4)(d-2)=0\\\\d= 4 \ or \ 2 \\\\Thus \ the \ four \ terms \ are:\\\\$ $\bf3,7,11,15\\\\OR\\\\-3,5,13,21$