Question 1016990
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an infinite geometric series with second term -8/9 and sum 2. what is the first term?
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a*r = {{{-8/9}}}    (1)    This is your first equation for the second term of GP.
                     Here "a"is the first term and "r" is the common ratio of the GP.

{{{a/(1-r)}}} = 2    (2)    This is your second equation for the sum.

From the first equation express r = {{{-8/(9a)}}} and substitute it into the second equation. You will get

{{{a/(1 - ((-8)/(9a))))}}} = 2,   or

{{{a}}} = {{{2*(1 + (8/(9a)))}}},   or

{{{a}}} = {{{2 + 16/(9a)}}}.

Now multiply both sides by 9a. You will get

{{{9a^2}}} = 18a + 16,   or 

{{{9a^2 - 18a - 16}}} = {{{0}}}.

Apply the quadratic formula to solve this quadratic equation. You will get

two roots {{{a[1]}}} = {{{8/3}}} and {{{a[2]}}} = -{{{2/3}}}.

The values of "r" that correspond to these values of "a" in accordance to (1), are

{{{r[1]}}} = {{{(-8/9)}}} : {{{8/3}}} = {{{-1/3}}}  and  {{{r[2]}}} = {{{(-8/9)}}} : {{{(-2/3)}}} = {{{4/3}}}.

The second {{{r[2]}}} = {{{4/3}}} has the modulus greater than 1 and therefore generates the "divergent" geometric progression. So, the second solution doesn't fit.

Now check the equality (2) for the first solution: {{{a[1]/(1-r[1])}}} = {{{8/3}}} : {{{(1-(-1/3))}}} = {{{8/3}}} : {{{(1 + 1/3)}}} = {{{8/3}}} : {{{4/3}}} = 2.  OK!

Thus there is a unique GP with the given second term and the given sum.

It is  {{{a[1]}}} =  {{{8/3}}},  {{{r[1]}}} = {{{-1/3}}}.
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