# to solve geometric progression

the battery in my watch is near the end of its useful life. Over the past 4 days it has lost 5 sec on day 1, a further 10 sec on day 2, a further 15 sec on day 3, and another 20 sec on day 4. After how many days will the watch have lost a total in excess of 5 min??

a1 = 5 = 5x1
a2 = 5 + 10 = 5x1 + (5x2)
a3 = 5 + 10 + 15 = 5x1 + (5x2) + (5x3)
a4 = 5 + 10 + 15 + 20 = 5x1 + (5x2) + (5x3) + (5x4)

an = 5 + 10 + 15 + 20 = 5x1 + (5x2) + (5x3) + (5x4) ......+ (5xn)

a = 5
how can i calculate the "r"??
S300 = a (r^n – 1)/(r-1)
pls help ??

Hint: it's not actually a geometric progression. A geometric progression is of the form $$a_1, a_1r, a_1r^2, a_1r^3, \dots$$ So for example, a geometric progression with $r = 5$ could look like $$1, 5, 25, 125, \dots$$ Notice that these are not the multiples of $5$, but powers, since the $n^\text{th}$ term is given by $ar^{n-1}$.
Now, what you have in your question is actually an arithmetic progression, which is of the form $$a_1, (a_1 + d), (a_1 + 2d), (a_1 + 3d), \dots$$ where in your case, $a_1 = 5$ and $d = 5$. (Check: $$\mathbf 5, 5 + 5 = \mathbf{10}, 5 + 2 \times 5 = \mathbf{15}, \dots)$$ A sum of arithmetic series is given by averaging the first and last terms, and multiplying by how many there are: $$S = \frac{n(a_1 + a_n)}{2}.$$ In our case, we don't know the last term: we need to find the number of terms before the sum exceeds a certain number. So we substitute (using the definition) $$a_n = a_1 + (n-1)d$$ to find $$S = \frac{n}{2}(2a_1 + (n-1)d).$$ With $a_1 = 5 = d$, we have $$S = \frac{n}{2}(10 + 5(n-1)).$$ Now see if you can use this to solve for the value of $n$ for which this sum exceeds 5 minutes.