Find the value of $4 + 5 + 6 + 7 + . . . . + 12.$

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Solution

The correct option is A 72
Given sum of numbers, 4+5+6+7+....+12 can be written as (1+2+3+4+....+12)-(1+2+3)

$∵$ Sum of first n natural numbers $= \frac{n (n + 1)}{2}$

$∴ 4 + 5 + 6 + 7 + . . . . + 12 = \frac{12 \times 13}{2} - \frac{3 \times 4}{2}$
$= 78 - 6 = 72$

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(\frac{1}{4} - \frac{1}{6})

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(\frac{2}{3} - \frac{5}{12}) \times (\frac{5}{8} - \frac{7}{12})

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

Q. Find the missing number.

$5$	$6$	$12$	$42$
$7$	$5$	$7$	$42$
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Q. Find the value of:

(1) 3⁶
(2) 6³
(3) 2⁶
(4) 1⁸
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Find the value of 4+5+6+7+....+12.

The correct option is A 72Given sum of numbers, 4+5+6+7+....+12 can be written as (1+2+3+4+....+12)-(1+2+3) ∵ Sum of first n natural numbers =n(n+1)2 ∴4+5+6+7+....+12=12×132−3×42 =78−6=72

Find the value of $4 + 5 + 6 + 7 + . . . . + 12.$

The correct option is A 72
Given sum of numbers, 4+5+6+7+....+12 can be written as (1+2+3+4+....+12)-(1+2+3)

$∵$ Sum of first n natural numbers $= \frac{n (n + 1)}{2}$

$∴ 4 + 5 + 6 + 7 + . . . . + 12 = \frac{12 \times 13}{2} - \frac{3 \times 4}{2}$
$= 78 - 6 = 72$