Question

(ii) Find four consecutive terms in an A.P. whose sum is 72 and
the ratio of the product of the 1st and the 4th terms to the
product of the 2nd and 3rd terms is 9 : 10.
(a) Let the four consecutive terms be a – 3d, a – d, a +d,
a + 3d.
(b) Using the first condition, find the value of a.
(c) Using the second condition, find the value of d and hence
the numbers.​

Answer 1

Answer:

(12,16,20,24) OR (24,20,16,12)

Step-by-step explanation:

Let the required terms be (a-3d),(a-d),(a+d) and (a=3d)

Now,

      ACCORDING TO THE QUESTION

a-3d+a-d+a+d+a+3d=72

4a=72

a=18

Also,

       $\frac{(a-3d)(a+3d) }{(a-3d)(a+3d)}=\frac{9}{10}$

        $\frac{a^{2}-9d^{2} }{a^{2}-d^{2} } =\frac{9}{10}$

       $10(a^{2} -9d^{2} )=9(a^{2} -d^{2} )$

      $10a^{2}-90d^{2} =9a^{2} -9d^{2}$

      $a^{2} =81d^{2}$

      $d^{2} =\frac{a^{2} }{81}$

      $d=\sqrt{\frac{a^{2} }{(9)^{2} } }$

     $d=\frac{18}{9}$

     d=±2

So, the required terms are (12,16,20,24) or (24,20,16,12)

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