finite, find the sum of 7 terms of 75,-15,3,-3/5

It appears the common ratio is -¹/₅

The sum of a geometric series is:

$${a}{\left({\frac{\left({\mathtt{1}}{\mathtt{\,-\,}}{{\mathtt{r}}}^{{\mathtt{n}}}\right)}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{r}}\right)}}\right)}$$

Where:

a = first term

r = common ratio

n = Number of terms of sequence

So the sum of the first 7 terms is:

75 * (1 - (-¹/₅)⁷) / (1 - -¹/₅)

= 75 * (1 - -0.0000128) / (1 + ¹/₅)

= 75 * 1.0000128 / 1.2

= 62.5008

 find the sum of 7 terms of 75,-15,3,-3/5

$$\small{ \text{ratio} = -\dfrac15 = -0.2 \qquad a_1 = 75 }\\\\ \small{ \text{geometric sequence: } a_n = a_1 \cdot r^{n-1} }\\\\ \begin{array}{rclclcr} \small{a_1 &=& 75 \cdot r^0 &=& 75 }\\ \\ \small{a_2 &=& 75 \cdot r^1 &=& 75 \cdot \left( -\dfrac15 \right)^1 &=& -15 }\\\\ \small{a_3 &=& 75 \cdot r^2 &=& 75 \cdot \left( -\dfrac15 \right)^2 &=& 3 }\\\\ \small{a_4 &=& 75 \cdot r^3 &=& 75 \cdot \left( -\dfrac15 \right)^3 &=& -0.6 }\\\\ \small{a_5 &=& 75 \cdot r^4 &=& 75 \cdot \left( -\dfrac15 \right)^4 &=& 0.12 }\\\\ \small{a_6 &=& 75 \cdot r^5 &=& 75 \cdot \left( -\dfrac15 \right)^5 &=& -0.024 }\\\\ \small{a_7 &=& 75 \cdot r^6 &=& 75 \cdot \left( -\dfrac15 \right)^6 &=& 0.0048 }\\\\ \end{array}\\\\ \small{ \mathrm{sum}_7 = 75-15+3-0.6+0.12-0.024+0.0048= 62.5008 }\\\\ \small{ \mathrm{sum}_7 = a_1 \cdot \left( \dfrac{ 1-r^7} {1-r} \right) = 75\cdot \left( \dfrac{ 1-(-0.2)^7} {1-(-0.2)} \right) =75\cdot \left( \dfrac{ 1+0.2^7} {1.2)} \right) =62.5008 }$$

$$\boxed{~~ \small{ \mathrm{sum}_n = a_1 \cdot \left( \dfrac{ 1-r^n} {1-r} \right) } ~~}$$