Sum of the series 96 + 24 + 6 + ....
Common ratio, r = \(\frac{24}{96}\) = \(\frac{1}{4}\)
First term, a = 96
Sum to infinity , S\(_∞\) = \(\frac{\text{a}}{1 - r}\) = \(\frac{96}{(1 - \frac{1}{4})}\)
S\(_∞\) = \(\frac{96}{\frac{3}{4}}\) = 96 x \(\frac{4}{3}\) = 32 x 4 = 128.
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