7th term of 1-2- 1-4- 1-6 - is

The correct option is D 1/14 We can see that the series has terms such that 12(1), 12(2), 12(3)..... So, the 7th term will be 12(7) = 114 ...

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Summation by Sigma Method

7th term of ...

Question

7th term of $\frac{1}{2}$ , $\frac{1}{4}$ , $\frac{1}{6}$ $\dots \dots$ is

\frac{1}{10}

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\frac{1}{12}

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\frac{1}{14}

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\frac{1}{16}

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Similar questions

Size of item	$4$	$6$	$8$	$10$	$12$	$14$	$16$
Frequency	$2$	$1$	$3$	$6$	$4$	$3$	$1$

:

$X$	$2 - 4$	$4 - 6$	$6 - 8$	$8 - 10$	$10 - 12$	$12 - 14$	$14 - 16$
$F$	$1$	$2$	$2$	$3$	$6$	$3$	$2$

{(\frac{1}{2})}^{- 1} + {(\frac{1}{3})}^{- 1} + {(\frac{1}{4})}^{- 1}

{(\frac{1}{2})}^{- 2} + {(\frac{1}{3})}^{- 2} + {(\frac{1}{4})}^{- 2}

Verify whether the given statement is true or false :

(i) $(\frac{5}{9} \div \frac{1}{3}) \div \frac{5}{2} = \frac{5}{9} \div (\frac{1}{3} \div \frac{5}{2})$

(ii) ${(- 16) \div \frac{6}{5}} \div \frac{- 9}{10} = (- 16) \div {\frac{6}{5} \div \frac{- 9}{10}}$

(iii) $(\frac{- 3}{5} \div \frac{- 12}{35}) \div \frac{1}{14} = \frac{- 3}{5} \div (\frac{- 12}{35} \div \frac{1}{14})$

{(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

{(1, 3), (1, 5), (2, 5)}

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