How do you find $a_1$ for each geometric series $S_n=33$ , $a_n=48$ , r=-2?

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Answer 1 · 2017-12-03T02:20:56

$a_1=3$

For geometric series, $S_n=a_1*(1-r^n)/(1-r)$ and $a_n=a_1*r^(n-1)$

Hence $a_1*[1-(-2)^n]/[1-(-2)]=33$ or $a_1*(1-r^n)=99$ and $a_1*(-2)^(n-1)=48$

After dividing first equation to second one,

$[a_1*(1-(-2)^n)]/[a_1*(-2)^(n-1)]=99/48$

$(1-(-2)^n)/(-2)^(n-1)=33/16$

$16*(1-(-2)^n)=33*(-2)^(n-1)$

$16-16*(-2)^n=-33/2*(-2)^n$

$-1/2*(-2)^n=16$

$(-2)^n=-32$ , so $n=5$

Thus,

$a_1*(-2)^(5-1)=48$

$16a_1=48$ , so $a_1=3$

How do you find a_1a_1 for each geometric series S_n=33S_n=33, a_n=48a_n=48, r=-2?