Question

If $S_n=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\cdots$ up to n terms

• A. $S_5=\frac{5}{16}$
• B. $S_{10}=\frac{10}{31}$
• C. $S_{n\to \infty }=\frac{1}{3}$
• D. $S_{2018}=\frac{2018}{2019}$

Answer 1

Answer: (A), (B), (C)

$S_{n\to \infty }=\frac{1}{3}$
$S_5=\frac{5}{16}$
$S_{10}=\frac{10}{31}$

$s_n=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+......$ upto $n$

We can see first part is $1,4,7$

Where $a=1$

$d=3$

$a_n=a+(n-1)d{a}_n=1+(n-1)3a_n=1+3n-3a_n=3n-2$

We can see ces part is $1,4,\mathrm{7....}$

$a=4$

$d=3$

$a_n=a+(n-1)d{a}_n=4+(n-1)3a_n=4+3n-3a_n=3n+1$

$a_n=\frac{1}{(3n+1)(3n-2)}$

$a_n=\frac{3}{3(3n+1)(3n-2)}{a}_n=\frac{1}{3}\left[\frac{3n+1-3n+2}{(3n+1)(3n-2)}\right]a_n=\frac{1}{3}[\frac{1}{3n-2}-\frac{1}{3n+1}]s_n=\frac{1}{3}{\sum }_{n=1}^n[\frac{1}{3n-2}-\frac{1}{3n+1}]s_n=\frac{1}{3}[1-\frac{1}{3n+1}]s_n=\frac{1}{3}\left[\frac{3n+1-1}{3n+1}\right]s_n=\frac{n}{3n+1}$

When $n=5s_5=\frac{5}{16}$

When $n=10$

$s_{10}=\frac{10}{31}$

When $n=\infty$

$s_{\infty }=\frac{1}{3}$

Also $s_{2018}\ne \frac{2018}{2019}$