Question 1199359
.
In a geometric sequence of real numbers, the sum of the first
two terms is 7 and the sum of the first six terms is 91. What is the sum of
the first four terms?
~~~~~~~~~~~~~~


<pre>
We are given

    {{{a[1]}}} + {{{a[2]}}} = 7    (1)

    {{{a[1]}}} + {{{a[2]}}} + {{{a[3]}}} + {{{a[4]}}} + {{{a[5]}}} + {{{a[6]}}} = 91      (2)   


In terms of "a" (the first term) and "r" (the common ratio) these equalities take the form

    a + ar = 7,       (3)

    a + ar + {{{ar^2}}} + {{{ar^3}}} + {{{ar^4}}} + {{{ar^5}}}  = 91    (4)


In (4), group the terms

    (a + ar) + ({{{ar^2}}} + {{{ar^3}}}) + ({{{ar^4}}} + {{{ar^5}}})  = 91.    (5)


Re-write (5) in an equivalent form

    (a + ar) + {{{r^2*(a+ar)}}} + {{{r^4(a+ar)}}}) = 91.    (6)


In (6), replace (a+ar) by the value of 7, based on (3).  You will get

    7 + {{{7r^2}}} + {{{7*r^4}}} = 91.   (7)


In (7), divide both sides by 7

    1 + {{{r^2}}} + {{{r^4}}} = 13,

or

    {{{r^4}}} + {{{r^2}}} - 12 = 0.    (8)


This biquadratic equation (8) is the quadratic equation relative {{{r^2}}}.  Solve it by factoring

    {{{(r^2+4)*(r^2-3)}}} = 0


Since we are given that the progression is in real numbers, {{{r^2+4}}}  can not be zero 
(it is positive at any real r), we conclude that only possible value of  {{{r^2}}}  is

    {{{r^2}}} = 3.


Then for the sum of the first four terms of this geometric progression  {{{S[4]}}}  we have
    {{{S[4]}}} = a + ar + {{{ar^2}}} + {{{ar^3}}} = (a + ar) + ({{{ar^2}}} + {{{ar^3}}}) = (a+ar) + {{{r^2(a+ar)}}}.


We substitute here  a+ar = 7 and {{{r^2}}} = 3, and we get
    {{{S[4]}}} = 7 + 3*7 = 7 + 21 = 28.


<U>ANSWER</U>.  The sum of the first 4 terms of this GP is 28.
</pre>

Solved.