If S_n= 16 11+ 111 16+ 116 21+ … to n terms, then 6S_n equals

Correct answer: B. (b) : S_n= 16 11+ 111 16+ 116 21+ …n^ th term = 15( 16- 111+ 111- 116+ … 15n+1- 15n+6) = 15( 16- 15n+6)= 15 ( 5n+6-66 (5n+6 ) )= n6 (5n+6 ) ⇒ 6S_n= n5n+6.

JEE Previous Year Question

$If S_{n} = \frac{1}{6 \cdot 11} + \frac{1}{11 \cdot 16} + \frac{1}{16 \cdot 21} + \dots to n terms, then 6 S_{n} equals$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

A $\frac{5 n - 4}{5 n + 6}$
B $\frac{n}{5 n + 6}$
C $\frac{2 n - 1}{5 n + 6}$
D $\frac{1}{5 n + 6}$

Correct answer: B

Solution

(b) : S_{n} = \frac{1}{6 \cdot 11} + \frac{1}{11 \cdot 16} + \frac{1}{16 \cdot 21} + \dots n^{th} term = \frac{1}{5} (\frac{1}{6} - \frac{1}{11} + \frac{1}{11} - \frac{1}{16} + \dots \frac{1}{5 n + 1} - \frac{1}{5 n + 6}) = \frac{1}{5} (\frac{1}{6} - \frac{1}{5 n + 6}) = \frac{1}{5} (\frac{5 n + 6 - 6}{6 ( 5 n + 6 )}) = \frac{n}{6 ( 5 n + 6 )} \Rightarrow 6 S_{n} = \frac{n}{5 n + 6} .

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