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There are a total of 4 terms. Since each term is twice the previous term, let 1st term=( a), 2nd term= (a*2), 3rd term=(a*2^2), then last term =( a*2^3). Stated that the last term is equal to 48, you can get an equation [48=(a*2^3)] to find a. The answer is a=6. We substitute it to the equation of terms above to find the numbers. .........The numbers would be 6,12,24 and 48.........Next,the formula for the sum would be "Sn=(a(1-r^n)/(1-r)" since the common ratio ( r )=2 and the first term ( a ) =6, we substitute it to the formula then we will get the sum (Sn)=90
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