Question 1150618
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The sum of the n terms of two arithmetic progression are in the ratio (3n+8):(7n+15). Find the ratio of their 12th terms
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<pre>
Our APs (arithmetic progressions) are

    a,  a+d,  a+2*d,  a+3*d, . . . 

    b,  b+e,  b+2*e,  b+3*e, . . . 

where  "a"  and  "b"  are first terms  and  "d"  and  "e"  are their common differences.   


In order for to answer the question, it is enough to know three ratios  {{{a/b}}},  {{{d/b}}}  and  {{{e/b}}}.

So, our goal now is to find these three ratios.



1)  At n= 1  we have


        {{{a/b}}} = {{{(3*1+8)/(7*1+15)}}} = {{{11/22}}} = {{{1/2}}}.         (1)  



2)  At n= 2  we have


        {{{(a+d)/(b+e)}}} = {{{(3*2+8)/(7*2+15)}}} = {{{14/29}}},  or

        29*(a+d) = 14*(b+e),

        29a + 29d - 14e = 14b.


    Divide both sides by "b" 

        29*(a/b) + 29*(d/b) - 14*(e/b) = 14.


    Substitute here {{{a/b}}} = {{{1/2}}} from (1) and multiply both sides by 2. You will get then

        58*(d/b) - 28*(e/b) = -1.       (2)



3)  At n= 3  we have


        {{{(a+2d)/(b+2e)}}} = {{{(3*3+8)/(7*3+15)}}} = {{{17/36}}},  or

        36*(a+2d) = 17*(b+2e),

        36a + 72d - 34e = 17b.


    Divide both sides by "b" 

        36*(a/b) + 72*(d/b) - 34*(e/b) = 17.


    Substitute here {{{a/b}}} = {{{1/2}}} from (1). You will get then

        72*(d/b) - 34*(e/b) = -1.       (3)



4)  Introduce new variables  D = {{{d/b}}}  and  E = {{{e/b}}}.  For these unknowns, from (2) and (3) you have this system of equations

        58*D - 28*E = -1        (2')

        72*D - 34*E = -1        (3')

    Solve it by any method you want / (you know).  The solution is  D = {{{3/22}}},  E = {{{7/22}}}.  Thus  {{{d/b}}} = {{{3/22}}},  {{{e/b}}} = {{{7/22}}}.



5)  Now we are in position to calculate  {{{a[12]/b[12]}}} = {{{(a+11*d)/(b+11*d)}}} = {{{((a/b) + 11*(d/b))/(1 + 11*(e/b))}}} = {{{(1/2 + 11*(3/22))/(1 + 11*(7/22))}}} = {{{((4/2))/((1+7/2))}}} = {{{4/9}}}.    <U>ANSWER</U>
</pre>

Solved.