Question

The sum of four consecutive num
sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first
was term to the product of two middle terms is 7: 15. Find the numbers.​

Answer 1

Answer:

(2,6,10,14) OR (14,10,6,2)

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=32

4a=32

a=8

Also,

       $\frac{(a-3d)(a+3d)}{(a-d)(a+d)} =\frac{7}{15}$

       $\frac{a^{2}-9d^{2} }{a^{2}-d^{2} } =\frac{7}{15}$

       $15(a^{2} -9d^{2} )=7(a^{2} -d^{2} )$

       $15a^{2} -135d^{2} =7a^{2} -7d^{2}$

       $8a^{2} =128d^{2}$

       $d^{2} =\frac{8a^{2} }{128}$

       $d^{2} =\frac{8(8)^{2} }{128}$

       $d=\sqrt{4}$

      d= ±2

S0, the required terms are (2,6,10,14) or (14,10,6,2)

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