Find the common difference of the following arithmetic progression.
$3 + 2 \sqrt{2}, 3 + 4 \sqrt{2}, 3 + 6 \sqrt{2}, 3 + 8 \sqrt{2}$

$3 + \sqrt{2}$

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$3$

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$\sqrt{2}$

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$2 \sqrt{2}$

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Solution

The correct option is D
$2 \sqrt{2}$

Common difference is the difference between two consecutive terms in an arithmetic sequence.
$(3 + 4 \sqrt{2}) - (3 + 2 \sqrt{2}) = 2 \sqrt{2}$
$(3 + 6 \sqrt{2}) - (3 + 4 \sqrt{2}) = 2 \sqrt{2}$
$(3 + 8 \sqrt{2}) - (3 + 6 \sqrt{2}) = 2 \sqrt{2}$

Hence, we can conclude that the common difference of the given AP is $2 \sqrt{2} .$

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Find the common difference of the following arithmetic progression. 3+2√2, 3+4√2, 3+6√2, 3+8√2

The correct option is D 2√2 Common difference is the difference between two consecutive terms in an arithmetic sequence. (3+4√2)−(3+2√2)=2√2 (3+6√2)−(3+4√2)=2√2 (3+8√2)−(3+6√2)=2√2 Hence, we can conclude that the common difference of the given AP is 2√2.

Find the common difference of the following arithmetic progression.
$3 + 2 \sqrt{2}, 3 + 4 \sqrt{2}, 3 + 6 \sqrt{2}, 3 + 8 \sqrt{2}$