Find the sum of the sequence 96 - 48 + 24 ……… upto 10 terms.

The above sequence is a G.P. with first term (a) = 96 and common ratio (r) = $\dfrac{-48}{96} = -\dfrac{1}{2}$.

By formula,

Sum of G.P. = $\dfrac{a(1 - r^n)}{1 - r}$

$\text{Sum upto 10 terms} = \dfrac{96 \times \Big[1 - \Big(-\dfrac{1}{2}\Big)^{10}\Big]}{1 - \Big(-\dfrac{1}{2}\Big)} \\[1em] = \dfrac{96 \times \Big[1 - \dfrac{1}{1024}\Big]}{1 + \dfrac{1}{2}} \\[1em] = \dfrac{96 \times \dfrac{1023}{1024}}{\dfrac{3}{2}} \\[1em] = \dfrac{12 \times \dfrac{1023}{128}}{\dfrac{3}{2}} \\[1em] = \dfrac{12 \times 1023 \times 2}{128 \times 3} \\[1em] = \dfrac{2046}{32} \\[1em] = 63\dfrac{15}{16}.$

Hence, sum of sequence = $63\dfrac{15}{16}.$