Question

1. $\sf \: if \: \: \dfrac{3 + 5 + 7 + . . . . + n \: terms }{5 + 8 + 11. . . . + 10 \: terms} = 7.then \: the \: value \: of \: n = ?$

2. $\sf \: if \: \: \: \dfrac{ a + bx}{a - bx} = \dfrac{b + cx}{b - cx} = \dfrac{c + dx}{c - dx} \: .then \: a , b , c , d \: are \: in$

Note:-

1. Irrelevant answers will be reported.

2. Answer will detailed explanation.

​

Answer 1

1. Consider the arithmetic series,

$\longrightarrow S_1=3+5+7+\,\dots\,n\ terms$

The n'th term of this series is,

$\longrightarrow a_n=3+(n-1)(5-3)$

$\longrightarrow a_n=3+2(n-1)$

$\longrightarrow a_n=2n+1$

So,

$\longrightarrow S_1=3+5+7+\,\dots\,+(2n+1)$

$\longrightarrow S_1=\dfrac{n}{2}\left[3+(2n+1)\right]$

$\longrightarrow S_1=n^2+2n$

Consider the arithmetic series,

$\longrightarrow S_1=5+8+11+\,\dots\,10\ terms$

The n'th term of this series is,

$\longrightarrow a_n=5+(n-1)(8-5)$

$\longrightarrow a_n=5+3(n-1)$

$\longrightarrow a_n=3n+2$

Then 10th term is,

$\longrightarrow a_{10}=3\times10+2$

$\longrightarrow a_{10}=32$

So,

$\longrightarrow S_1=5+8+11+\,\dots\,+32$

$\longrightarrow S_1=\dfrac{10}{2}\left[5+32\right]$

$\longrightarrow S_1=185$

Now,

$\longrightarrow\dfrac{3+5+7+\,\dots\,n\ terms}{5+8+11+\,\dots\,10\ terms}=7$

$\longrightarrow\dfrac{n^2+2n}{185}=7$

$\longrightarrow n^2+2n-1295=0$

$\longrightarrow n^2+37n-35n-1295=0$

$\longrightarrow n(n+37)-35(n+37)=0$

$\longrightarrow(n+37)(n-35)=0$

Since $n>0,$

$\longrightarrow\underline{\underline{n=35}}$

2. Consider,

$\longrightarrow\dfrac{a+bx}{a-bx}=\dfrac{b+cx}{b-cx}$

By rule of componendo and dividendo, we get,

$\longrightarrow\dfrac{bx}{a}=\dfrac{cx}{b}$

$\longrightarrow\dfrac{b}{a}=\dfrac{c}{b}$

Similarly, consider,

$\longrightarrow\dfrac{b+cx}{b-cx}=\dfrac{c+dx}{c-dx}$

Then we get,

$\longrightarrow\dfrac{c}{b}=\dfrac{d}{c}$

Now,

$\longrightarrow\dfrac{b}{a}=\dfrac{c}{b}=\dfrac{d}{c}$

This implies $a,\ b,\ c,\ d$ are in Geometric Progression.