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Mathematics
  1. The first and fifth terms of a linear sequence (A.P.) are 2 and -50 respectively. Find the sum of first 2 1 21 terms.
    A. -5229
    B. -3614
    C. -2688
    D. -1344
  2. Given that r = ( 12 N , 300 ) r=(12 N,300^(@)), express r r in the form ( p q ) ([p],[],[q]).
    A. ( 6 3 6 ) N quad([6sqrt3],[],[6])N
    B. ( 12 3 12 ) N quad([-12sqrt3],[],[12])N
    C. ( 6 3 6 ) N quad([-6sqrt3],[],[6])N
    D. ( 6 6 3 ) N quad([6],[],[-6sqrt3])N
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  1. Let d d be the common difference of the arithmetic progression. Then the fifth term is 2 + 4 d = 50 2+4d=-50, so 4 d = 52 4d=-52 and d = 13 d=-13.
    The sum of the first 21 terms is given by:
S 21 = 21 2 ( 2 + ( 21 1 ) ( 13 ) ) = 21 2 ( 266 ) = (A) 5229 S_(21)=(21)/(2)(2+(21-1)(-13))=(21)/(2)(-266)=(A)-5229
Therefore, the sum of the first 21 terms is 5229 -5229.
14. We can convert the given polar coordinates to rectangular coordinates by using the formulas:
x = r cos θ and y = r sin θ x=r cos thetaquad"and"quad y=r sin theta
Substituting the given values, we get:
x = 12 N cos 300 = 12 N 3 2 = 6 3 N x=12 N cos 300^(@)=12 N*(sqrt3)/(2)=6sqrt3N
y = 12 N sin 300 = 12 N ( 1 2 ) = 6 N y=12 N sin 300^(@)=12 N*(-(1)/(2))=-6N
Therefore, r = x 2 + y 2 = ( 6 3 N ) 2 + ( 6 N ) 2 = 6 3 N r=sqrt(x^(2)+y^(2))=sqrt((6sqrt3N)^(2)+(-6N)^(2))=6sqrt3N, and the direction of r r is given by tan 1 ( y / x ) = tan 1 ( 3 / 3 ) tan^(-1)(y//x)=tan^(-1)(-sqrt3//3). Note that this angle is in the third quadrant, which has a negative y y-coordinate.
Therefore, r = (A) ( 6 3 6 ) N r=(A)([6sqrt3],[],[6])N.
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