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Algebra Level 1

Find the sum of the geometric progression 2 3 , 1 , 3 2 , . . . \frac{2}{3}, -1,\frac{3}{2},... up to 7 7 terms.

95 464 \dfrac{95}{464} 563 191 \dfrac{563}{191} 463 96 \dfrac{463}{96} 465 93 \dfrac{465}{93}

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1 solution

Relevant wiki: Geometric Progression Sum

The common ratio is r = 1 2 3 = 3 2 1 = 3 2 r=\dfrac{-1}{\dfrac{2}{3}}=\dfrac{\dfrac{3}{2}}{-1}=\dfrac{-3}{2} .

The sum of the terms of a geometric progression is given by s = a 1 a 1 r n 1 r s=\dfrac{a_1-a_1r^n}{1-r} where a 1 a_1 is the first term, r r is the common ratio, and n n is the number of terms. Substituting, we have

s = 2 3 2 3 ( 3 2 ) 7 1 + 3 2 = 463 96 s=\dfrac{\dfrac{2}{3}-\dfrac{2}{3}\left(\dfrac{-3}{2}\right)^7}{1+\dfrac{3}{2}}=\boxed{\dfrac{463}{96}}

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