How do you find the sum of the infinite geometric series -13and1/2 + 9 – 6 + . . . ?

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Answer 1 · 2016-04-17T08:30:22

Sum of the series is $- 8.10$ $-8.10$

As the series is ${-13 1/2,+9,-6,.......}$

its first term is $-13 1/2=-27/2$ and it is a geometric series as

$(+9)/(-27/2)=-(9xx2)/27=-2/3$ and also $-6/9=-2/3$

Now sum of a geometric series upto $n^(th)$ term, with first term as $a$ and ratio as $r$ is given by

$a(r^n-1)/(r-1)$ if $|r|>1$ and $a(1-r^n)/(1-r)$ if $|r|<1$

If $n->oo$ the latter i.e. when $|r|<1$ is $a/(1-r)$

Hence sum of the series is $(-27/2)/(1-(-2/3))=(-27/2)/(1+2/3)$

= $(-27/2)/(5/3)=-27/2xx3/5=-81/10=-8.10$