Solve the linear equation- n2 - 3n4 - 5n6 - 21

The correct option is D n = 36 Given, n 2 3n 4 + 5n 6 =21 ⟹ 6n 9n+10n 12 =21 #160; #160; #160; #160;... [Taking LCM on LH ...

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Standard XII

Mathematics

Absolute Value Function

Solve the lin...

Question

Solve the linear equation: $\frac{n}{2} - \frac{3 n}{4} + \frac{5 n}{6} = 21$

n = 22

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n = 33

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n = 29

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n = 36

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Solution

The correct option is D $n = 36$
Given, $\frac{n}{2} - \frac{3 n}{4} + \frac{5 n}{6} = 21$
$⟹ \frac{6 n - 9 n + 10 n}{12} = 21$ ... [Taking LCM on LHS]
$⟹ \frac{7 n}{12} = 21$
$⟹ \frac{n}{12} = 3$ .... [Divide both sides by 7]
$⟹ n = 3 \times 12$ ...[Use cross-multiplication]
$⟹ n = 36$

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Similar questions

Solve linear equation:
$\frac{n}{2} - \frac{3 n}{4} + \frac{5 n}{6} = 21$

Q. Question 2

\frac{n}{2} - \frac{3 n}{4} + \frac{5 n}{6} = 21

Q. Solve

\frac{2}{3} (n + 6) - \frac{1}{5} (n - 4) = \frac{3}{7} (n + 12)

, then

n =

1^{6} - 2^{6} + 3^{6} - . . . . + (- 1)^{n - 1} n^{6} = \frac{(- 1)^{n - 1}}{2} (n^{6} + 3 n^{5} - 5 n^{3} + 3 n)

Find $(3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + . . . .$ to 10 terms.

Show that $\frac{1 \times 2^{2} + 2 \times 3^{2} + . . . . + n \times (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + . . . . + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}$

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Solve the linear equation:n2−3n4+5n6=21

The correct option is D n=36Given, n2−3n4+5n6=21⟹6n−9n+10n12=21 ... [Taking LCM on LHS]⟹7n12=21⟹n12=3 .... [Divide both sides by 7]⟹n=3×12 ...[Use cross-multiplication] ⟹n=36

Solve the linear equation: $\frac{n}{2} - \frac{3 n}{4} + \frac{5 n}{6} = 21$

The correct option is D $n = 36$
Given, $\frac{n}{2} - \frac{3 n}{4} + \frac{5 n}{6} = 21$
$⟹ \frac{6 n - 9 n + 10 n}{12} = 21$ ... [Taking LCM on LHS]
$⟹ \frac{7 n}{12} = 21$
$⟹ \frac{n}{12} = 3$ .... [Divide both sides by 7]
$⟹ n = 3 \times 12$ ...[Use cross-multiplication]
$⟹ n = 36$