Question 1140428
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Given:
{{{a = 3}}}
{{{ar^(n-1) = 768}}}<br>
So
{{{r^(n-1) = 768/3 = 256 = 4^4 = 2^8}}}<br>
So the series might have a common ratio of 4 with 5 terms, or a common ratio of 2 with 9 terms.<br>
If the series is 5 terms with a common ratio of 4, the sum would be<br>
{{{a(r^n-1)/(r-1) = 3(4^5-1)/(4-1) = 3(1023)/3 = 1023}}}<br>
not the right sum...<br>
If the series is 9 terms with a common ratio of 2, the sum would be<br>
{{{a(r^n-1)/(r-1) = 3(2^9-1)/(2-1) = 3(511)/1 = 1533}}}<br>
That's the right sum.<br>
ANSWER: common ratio 2; number of terms 9