If Sn=(1/6.11)+(1/11.16)+(1/16.21)+.... to n terms, then 6Sn equals

Question

If Sn=(1/6.11)+(1/11.16)+(1/16.21)+.... to n terms, then 6Sn equals (A)  (5n-4/5n+6)  (B)  (n/(5n+6))  (C)  (2n-1/5n+6)  (D)  (1/(5n+6))

Tardigrade Admin · Accepted Answer

Given that, Sn=(1/6.11)+(1/11.16)+(1/16.21)+....+n terms =(1/5)( (1/6)-(1/11)+(1/11)-(1/16)+....(1/5n+1)-(1/5n+6) ) =(1/5)( (1/6)-(1/11)+(1/11)-(1/16)+....+(1/5n+1)-(1/5n+6) ) =(1/5)( (1/6)-(1/5n+6) ) =(1/5)( (5n+6-6/6(5n+6)) ) =(n/6(5n+6)) ⇒ 6Sn=(n/5n+6)

Q. If $S_{n} = \frac{1}{6.11} + \frac{1}{11.16} + \frac{1}{16.21} + ....$ to $n$ terms, then $6 S_{n}$ equals

$\frac{5 n - 4}{5 n + 6}$

$\frac{n}{( 5 n + 6 )}$

$\frac{2 n - 1}{5 n + 6}$

$\frac{1}{( 5 n + 6 )}$

Q. If Sn​=6.111​+11.161​+16.211​+.... to n terms, then 6Sn​ equals

5n+65n−4​

(5n+6)n​

5n+62n−1​

(5n+6)1​

Given that, Sn​=6.111​+11.161​+16.211​+....+n terms =51​(61​−111​+111​−161​+....5n+11​−5n+61​) =51​(61​−111​+111​−161​+....+5n+11​−5n+61​) =51​(61​−5n+61​) =51​(6(5n+6)5n+6−6​) =6(5n+6)n​ ⇒ 6Sn​=5n+6n​

Q. If $S_{n} = \frac{1}{6.11} + \frac{1}{11.16} + \frac{1}{16.21} + ....$ to $n$ terms, then $6 S_{n}$ equals

$\frac{5 n - 4}{5 n + 6}$

$\frac{n}{( 5 n + 6 )}$

$\frac{2 n - 1}{5 n + 6}$

$\frac{1}{( 5 n + 6 )}$