Question
Mathematics
1. The first and fifth terms of a linear sequence (A.P.) are 2 and -50 respectively. Find the sum of first $21$21 terms.
A. -5229
B. -3614
C. -2688
D. -1344
2. Given that $r=\left(12N,{300}^{\circ }\right)$r=(12 N,300^(@)), express $r$r in the form $\left(\begin{array}{l}p\\ \\ q\end{array}\right)$([p],[],[q]).
A. $\phantom{\rule{1em}{0ex}}\left(\begin{array}{c}6\sqrt{3}\\ \\ 6\end{array}\right)N$quad([6sqrt3],[],[6])N
B. $\phantom{\rule{1em}{0ex}}\left(\begin{array}{c}-12\sqrt{3}\\ \\ 12\end{array}\right)N$quad([-12sqrt3],[],[12])N
C. $\phantom{\rule{1em}{0ex}}\left(\begin{array}{c}-6\sqrt{3}\\ \\ 6\end{array}\right)N$quad([-6sqrt3],[],[6])N
D. $\phantom{\rule{1em}{0ex}}\left(\begin{array}{c}6\\ \\ -6\sqrt{3}\end{array}\right)N$quad([6],[],[-6sqrt3])N
Solve problem with AI
1. Let $d$d be the common difference of the arithmetic progression. Then the fifth term is $2+4d=-50$2+4d=-50, so $4d=-52$4d=-52 and $d=-13$d=-13.
The sum of the first 21 terms is given by:
${S}_{21}=\frac{21}{2}\left(2+\left(21-1\right)\left(-13\right)\right)=\frac{21}{2}\left(-266\right)=\overline{)\mathbf{\text{(A)}}\phantom{\rule{0ex}{0ex}}-5229}$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:
$x=r\mathrm{cos}\theta \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}\theta$x=r cos thetaquad"and"quad y=r sin theta
Substituting the given values, we get:
$x=12N\mathrm{cos}{300}^{\circ }=12N\cdot \frac{\sqrt{3}}{2}=6\sqrt{3}N$x=12 N cos 300^(@)=12 N*(sqrt3)/(2)=6sqrt3N
$y=12N\mathrm{sin}{300}^{\circ }=12N\cdot \left(-\frac{1}{2}\right)=-6N$y=12 N sin 300^(@)=12 N*(-(1)/(2))=-6N
Therefore, $r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{\left(6\sqrt{3}N{\right)}^{2}+\left(-6N{\right)}^{2}}=6\sqrt{3}N$r=sqrt(x^(2)+y^(2))=sqrt((6sqrt3N)^(2)+(-6N)^(2))=6sqrt3N, and the direction of $r$r is given by ${\mathrm{tan}}^{-1}\left(y/x\right)={\mathrm{tan}}^{-1}\left(-\sqrt{3}/3\right)$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, $r=\overline{)\mathbf{\text{(A)}}\phantom{\rule{0ex}{0ex}}\left(\begin{array}{c}6\sqrt{3}\\ \\ 6\end{array}\right)N}$r=(A)([6sqrt3],[],[6])N.
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