Choose the correct answer from Column II for the questions in Column I.

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Solution

A. $\sum n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

B. Series is $\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, . . . . . .$
$a = \frac{3}{2}, r = \frac{1}{2}$
$t_{10} = a r^{9} = \frac{3}{2} \times \frac{1}{2^{9}} = \frac{3}{1024}$

C. $lim x \to 3 \frac{x^{3} - 27}{x - 3} = lim x \to 3 \frac{x^{3} - 3^{3}}{x - 3} = 3 \times 3^{3 - 1} = 27$

D.If $x > 0, y < 0,$ the $(x, y)$ lies in the $I V$ quadrant.

E. $| x | \leq a =$
$x \leq a$
and $- x \leq a \Rightarrow x \geq - a$
$∴ - a \leq x \leq a$

$A - (4), B - (1), C - (7), D - (2), E - (3)$

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Match column I with column II and choose the correction answer

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Choose the correct answer from Column II for the questions in Column I.

A. ∑n2=n(n+1)(2n+1)6B. Series is 32,34,38,......a=32,r=12t10=ar9=32×129=31024C. limx→3x3−27x−3=limx→3x3−33x−3=3×33−1=27D.If x>0,y<0, the (x,y) lies in the IV quadrant.E. |x|≤a=x≤aand −x≤a⇒x≥−a∴−a≤x≤aA−(4),B−(1),C−(7),D−(2),E−(3)