Question 1148035
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The sum of the first three terms is 37/8:
{{{a+ar+ar^2 = 37/8}}}<br>
The sum of the first six terms is 3367/512:
{{{a+ar+ar^2+ar^3+ar^4+ar^5 = 3367/512}}}<br>
The sum of terms 4 to 6 is the difference, 3367/512-37/8:
{{{ar^3+ar^4+ar^5 = 3367/512-37/8 = 3367/512-2368/512 = 999/512}}}<br>
The sum of terms 4 to 6 is the sum of the first three terms, multiplied by r^3.  That is,
{{{ar^3+ar^4+ar^5 = r^3(a+ar+ar^2)}}}
{{{999/512 = r^3(37/8)}}}
{{{r^3 = (999/512)*(8/37) = 27/64}}}
{{{r = 3/4}}}<br>
The sum of the first three terms is 37/8:
{{{a+(3/4)a+(9/16)a = 37/8}}}
{{{(37/16)a = 37/8}}}
{{{a = 2}}}<br>
ANSWERS: first term 2; common ratio 3/4