Question

If $\frac{1+2+3+4+... \rm upto \ n \ terms}{1.2+2.3+3.4+... \rm upto \ n \ terms} = 0.03$, find n.

Answer 1

Solution:

If $\frac{1+2+3+4+... \rm upto \ n \ terms}{1.2+2.3+3.4+... \rm upto \ n \ terms} = 0.03$

to find n,apply sum of n terms of AP in the numerator and denominator

$1 + 2 + 3 + 4...upto \: n \: terms \\ \\ here \: a = 1 \\ \\ d = 1 \\ \\ S_{n} = \frac{n}{2} [2a + (n - 1)d] \\ \\ S_{n}= \frac{n}{2} (2 + (n - 1)1) \\ \\ = \frac{n}{2} (2 + n - 1) \\ \\S_{n}= \frac{n}{2} (1 + n)$
for denominator

$1.2 + 2.3 + 3.4 + ...up \: to \: n \: terms \\ \\ here \: a = 1.2 \\ \\ d = 1.1\\ \\ S_{n} = \frac{n}{2} (2.4 + (n - 1)1.1) \\ \\ = \frac{n}{2} (2.4 + (n - 1)1.1) \\ \\S_{n}= \frac{n}{2} (2.4 + 1.1n - 1.1) \\ \\ S_{n}= \frac{n}{2} (1.3 + 1.1n)...eq2$

According to the question

$\frac{ \frac{n}{2} (1 + n)}{ \frac{n}{2}(1.3 + 1.1n) } = 0.03 \\ \\ 1 + n = 0.03(1.3 + 1.1n) \\ \\ 1 + n = 0.039 + 0.033n \\ \\ n - 0.033n = 0.039 - 1 \\ \\ - 0.967n = 0.961 \\ \\ n = \frac{ - 0.961}{0.967} \\ \\ n = - 0.99$
n can not be negative and fractional.

*please check the data you have provided,there might some mistake while typing.

Hope it helps you