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Q : 4       Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

              (v)  \small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},... 

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Given series is
\small 3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...
Now,
first term to this series is = 3
Now,
a_1 = 3 \ \ and \ \ a_2 = 3+\sqrt2 \ \ and \ \ a_3 = 3+2\sqrt2 \ \ and \ \ a_4 = 3+3\sqrt2
a_2-a_1 = 3+\sqrt2-3= \sqrt2
a_3-a_2 = 3+2\sqrt2-3-\sqrt2 = \sqrt2
a_4-a_3 = 3+3\sqrt2-3-2\sqrt2 = \sqrt2
We can clearly see that difference between terms are equal  and equal to \sqrt2
Hence, given series is in AP
Now, next three terms are
a_5=a_4+d = 3+3\sqrt2+\sqrt2=3+4\sqrt2
a_6=a_5+d = 3+4\sqrt2+\sqrt2=3+5\sqrt2
a_7=a_6+d = 3+5\sqrt2+\sqrt2=3+6\sqrt2
Therefore, next three terms of given series are  3+4\sqrt2, 3+5\sqrt2,3+6\sqrt2

Posted by

Gautam harsolia

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