Question 1205956
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The third term of a g.p is 12 and the first term is 48. find the sum of the first 11 terms
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        This problem allows two solutions :   there are two  GPs  satisfying the given condition,
        and they produce two different sums,  i.e. two different answers.


        Thus the solution by @mananth is  INCOMPLETE  and,  therefore,  partly  INCORRECT.


        I came to bring a correct solution.



<pre>
For the third term of this GP, we have this equation

    12 = 48*r^2.


It gives

    r^2 = {{{12/48}}} = {{{1/4}}},

    r = {{{sqrt(1/4)}}} = +/- {{{1/2}}}.


So, there are two values for common ratio: one is  r = 1/2; the other is r = -1/2.


With common ratio 1/2, the sum of the first 11 terms of the GP is 

    {{{S[11]}}} = {{{48*(1 - (1/2)^11)/(1-1/2)}}} = {{{48*((2^11-1)/2^11)/((1/2))}}} = {{{48*2*((2^11-1)/2^11)}}} = {{{3*(2047/64)}}} = 95.953125  (rounded)


With common ratio -1/2, the sum of the first 11 terms of the GP is 


    {{{S[11]}}} = {{{48*(1 + (1/2)^11)/(1+1/2)}}} = {{{48*((2^11+1)/2^11)/((3/2))}}} = {{{48*2*(2^11+1)/(3*2^11)}}} = {{{32*(2049/2^11)}}} = {{{2049/64}}} = 32.015625  (rounded).


<U>ANSWER</U>. Two possible answers are  {{{3*(2047/64)}}} = 95.953125 (with r = 1/2)  and  {{{2049/64}}} = 32.015625  (with r = -1/2).
</pre>

Solved.