2015/062) A problem in AP

There are 2 sets of numbers each consisting of 3 terms in A.P & sum of each set is 15. The common difference of the $1^{st}$ set is greater than the common difference of the $2^{nd}$ set by 1 and the ratio of product of the $1^{st}$ set is to the product of $2^{nd}$ set is 7 to 8.Find the numbers. 

Solution
 
As sum of 3 terms is 15 so middle term is 5


let the common difference in $1^{st}$ set be a, so common difference in $2^{nd}$ set is (a-1)

then numbers in 1^st set are 5-a, 5, 5+ a and in 2^nd set are the numbers are 6-a, 5, 4+ a

now as per given condition $\dfrac{(5-a)5(5+a)}{((6-a) 5 (4+a))} = \dfrac{7}{8}$

or $8(25-a^2)= 7(6-a)(4+a) = 7(24+ 2a - a^2)$

or $200- 8a^2 = 168 + 14a - 7a^2$ or $a^2 + 14 a - 32 = 0$

$(a-2)(a+16) = 0$

a= 2 gives a- 1 = 1 gives 1st series = 3,5,7 and second series = 4,5,6

a = - 16 gives a- 1= - 17 the 1st series = 21,5, -11 and second series = 22,5, - 12.

refer to http://in.answers.yahoo.com/question/index;_ylt=Ar2kcwS_d26mHg5f2W.h9F.RHQx.;_ylv=3?qid=20130222042246AAW7rSk