WAEC Further Mathematics Past Questions & Answers - Page 38

ar = \(\frac{2}{9}\) .....(i) 

ar\(^3\) = \(\frac{8}{81}\) ......(ii) 

\(\frac{ar3}{ar} = \frac{8}{81} \times \frac{9}{2}\) 

r\(^2 = \frac{4}{9}\) 

r = \(\sqrt{\frac{4}{9}}\) 

= \(\frac{2}{3}\) 

ar = \(\frac{2}{9}\) 

a(\(\frac{2}{3}\)) = \(\frac{2}{9}\)

a = (\(\frac{2}{3}\)) = \(\frac{2}{9}\)

a = \(\frac{2}{9} \times \frac{3}{2}\)

a = \(\frac{1}{3}\) 

T\(_r\) = ar\(^5\) = (\(\frac{1}{3}\))(\(\frac{2}{5}\))\(^5\) 

= \(\frac{32}{729}\) 

PQ = \(\begin {pmatrix} 2 & 3\\ -4 & 1\end {pmatrix}\) = \(\begin {pmatrix} 6 \\ 8 \end {pmatrix}\) = \(\begin {pmatrix} 36 \\ - 16\end {pmatrix}\)

\(\begin {pmatrix} 36 \\ -16 \end {pmatrix} \) = k = \(\begin {pmatrix} 45 \\ -20 \end {pmatrix} \)

k = \(\frac{36}{45}\) 

= \(\frac{4}{5}\)

distance = \(\sqrt{(3 - (-5)^2 + (-1 - (-2)^2)}\)

= \(\sqrt{8^2 + 1^2}\)

= \(\sqrt{65}\) units

\(\sqrt{4x^2 + 1}\) = \(\frac{13x}{6}\)

4x\(^2\) + 1 = \(\frac{169x^2}{36}\)

4 + x\(^2\)  = \(\frac{169x^2}{36}\) 

cross multiply

169x\(^2\) - 144x\(^2\) = 36

25x\(^2\) = 36

x\(^2\) = \(\frac{36}{25}\)

: x = \(\pm\frac{6}{5}\)

d = -3 - (-7) = 4

S\(_{20}\) = \(\frac{20}{2}\){2(-7) + (20 - 1) 4}

= 10(- 14 + 76)

= 620