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` `There are 2 sets of numbers each consisting of 3 terms in AP and the sum to each set is 15. the common difference of the first set is greater by 1 than the common difference of the second set. The product of the first set is to the product of the second set as 7 to 8.Find the numbers.

7 years ago

Dear sudeep

let the number is first set

a-d, a,a+d

amd number in the second set

b-(d-1) , b , b+(d-1)

give (a-d) + a + (a+d) =15

3a =15

a =5

and b-(d-1) + b + b+(d-1) =15

3b =15

b =5

and given (a-d)a(a+d)/{b-(d-1)}b{b+(d-1)} = 7/8

8[a^2-d^2]=7[a^2 -(d-1)^2]

solve it u will get the answer

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Badiuddin

7 years ago

Dear Sudeep,

first set: three terms are a-d, a, a+d

a-d + a + a+d = 15

a = 5, therefore terms are 5-d, 5, 5+d

second set: three terms are a-d+1, a, a+d-1

a = 5, terms are 6-d, 5, 4+d

product of first set = (5-d)(5)(5+d)

product of second set = (6-d)(5)(4+d)

(5-d)(5)(5+d)/(6-d)(5)(4+d) = 7/8

d

^{2}-25 / d^{2}- 2d - 24 = 7/8d

^{2}+ 14d - 32 = 0(d+16)(d-2) = 0

for d = 2, first set: 3, 5, 7

second set: 4, 5, 6

for d = -16

first set : 21, 5, -11

second set: 22, 5, -12

Please feel free to post as many doubts on our discussion forum as you can. If you find any question

Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We

are all IITians and here to help you in your IIT JEE preparation.

All the best Sudeep !!!

Regards,

Askiitians Experts

Gaurav Aggarwal

7 years ago

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