Question 1076783
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<pre>
Let "a" be the middle term of the AP, and "d" be the common difference.
Then you have these equations

{{{(a-2d)^2 + (a-d)^2 + a^2 + (a+d)^2 + (a+2d)^2}}} = {{{20*a^2}}},    (1)    and

{{{(a-2d)*(a-d)*a*(a+d)*(a+2d)}}} = {{{80}}}.             (2)


Simplify (1) and (2):

{{{5a^2 + 10d^2}}} = {{{20*a^2}}}          (1')     and
{{{a*(a^2-d^2)*(a^2-4d^2)}}} = {{{80}}}    (2')


Simplify (1'):

{{{a^2 + 2d^2}}} = {{{4a^2}}},   which is the same as  {{{3a^2}}} = {{{2d^2}}},  which is the same as

{{{d^2}}} = {{{(3/2)*a^2}}}.               (1'')


Then (2') becomes (after substitution (1'') )

{{{a*(a^2 - (3/2)*a^2)*(a^2 - 4*(3/2)*a^2)}}} = {{{80}}},   or

{{{a*((-1/2)*a^2)*(-5a^2)}}} = {{{80}}},   or

{{{a^5}}} = {{{(80*2)/5}}} = {{{32}}}.


Then a = {{{root(5,32)}}} = 2.


<U>Answer</U>.  The middle term under the question is 2.
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Solved.