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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.

                (xiii)  \small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...

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Given series is
\small \sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},...
Now,
the first term to this series is = \sqrt3
Now,
a_1 = \sqrt3 \ \ and \ \ a_2 = \sqrt6 \ \ and \ \ a_3 = \sqrt9 \ \ and \ \ a_4 = \sqrt{12}
a_2-a_1 = \sqrt6-\sqrt3 =\sqrt3(\sqrt2-1)
a_3-a_2 = 3-\sqrt3 =\sqrt3(\sqrt3-1)
We can clearly see that the difference between terms are not equal  
Hence, given series is not in AP
 

Posted by

Gautam harsolia

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