Question

the sum and product of three consecutive terms of an AP are 12 and 48 respectively find 3 terms​

Answer 1

Answer:

3 terms of AP = (2, 4, 6)  or (6, 4, 2)

Step-by-step explanation:

We are given that the sum and product of three consecutive terms of an AP are 12 and 48 respectively.

Let three consecutive terms of AP be a, a+d, a+2*d , where a = first term and d = common difference .

Sum of three consecutive terms of an AP is 12 i.e.;

    a + (a+d) + (a+2*d) = 12

     3*a + 3*d = 12

        a + d = 4

           d = 4 - a -------- [Equation 1]

Also, product of three consecutive terms of an AP is 48 i.e;

     a * (a+d) * (a+2*d) = 48

     a * 4 * (a + 2(4-a)) = 48   { Using equation 1}

        4a * (a + 8 - 2a) = 48

         4a * (8 - a) = 48

          $4a^{2} -32a +48 = 0$

           $a^{2} -8a +12 = 0$  

            $a^{2} -6a -2a+12 = 0$

            a(a - 6) - 2(a - 6) = 0

So, either (a - 2) = 0 or  (a - 6) = 0

Hence, a = 2 or a = 6

If a = 2 ,then d = 4 - 2 = 2    and   If a = 6, then d = 4 - 6 = -2

• If a = 2 and d = 2, then first 3 terms of AP = 2, 4, 6 .
• If a = 6 and d = -2, then first 3 terms of AP = 6, 4, 2 .