Question 1205015
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The 5th term of a GP is 8, the third term is 4, and the sum of the first ten terms is positive. 
Find the first term, the common ratio, and the sum of the first ten {{{highlight(cross(years))}}} terms.
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<pre>
    {{{a[5]}}} = 8

    {{{a[3]}}} = 4

implies

    {{{a[1]*r^4}}} = 8    (1)

    {{{a[1]*r^2}}} = 4    (2)


Divide equation (1) by equation (2)  (both sides).  You will get

    {{{r^2}}} = 2,  r = +/- {{{sqrt(2)}}}.


Let's consider case  r = {{{sqrt(2)}}}  first  (positive value).  We then have

    {{{a[1]}}} = {{{4/r^2}}} = {{{4/(sqrt(2))^2}}} = {{{4/2}}} = 2.

    {{{S[10]}}} = {{{a[1]*((r^10-1)/(r-1))}}} = {{{2*((sqrt(2)^10-1)/(sqrt(2)-1))}}} = {{{2*((2^5-1)/(sqrt(2)-1))}}} = {{{2*((32-1)/(sqrt(2)-1))}}} = {{{62/(sqrt(2)-1)}}} = {{{62*(sqrt(2)+1)}}}.


Now, if  r = {{{-sqrt(2))}}}, then the sum  {{{S[10]}}} will be, OBVIOUSLY, negative; so, due to the condition, we reject this case.


<U>ANSWER</U>.  {{{a[1]}}} = 2;  r = {{{sqrt(2)}}},  {{{S[10]}}} = {{{62*(sqrt(2)+1)}}}.
</pre>

Solved.


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Sorry, by mistake, I overlayed the post by @MathLover1 by my post . . . 


I regret very much, sorry . . .