(a) The 3rd and 8th terms of an arithmetic progression (A.P) are -9 and 26 respectively. Find the : (i) common difference ; (ii) first term.
(b)
In the diagram (overline{PQ} || overline{YZ}), |XP| = 2cm, |PY| = 3 cm, |PQ| = 6 cm and the area of (Delta) XPQ = 24(cm^{2}).Calculate the area of the trapezium PQZY.
Explanation
(a) (T_{n} = a + (n – 1)d) (terms of an A.P)
3rd term = -9 ; 8th term = 26
(implies T_{3} = a + 2d = -9 … (1))
(implies T_{8} = a + 7d = 26 ….. (2))
(i) Solving (2) – (1), we have
(7d – 2d = 26 – (-9))
(5d = 35 implies d = 7)
(ii) Putting d = 7 in (1) above, we have
(a + 2(7) = -9)
(a + 14 = -9)
(a = – 9 – 14 = -23)
(b) (frac{XP}{XY} = frac{2}{5}) (similar triangles)
(frac{24}{text{Area of Delta XYZ} = frac{2^{2}}{5^{2}})
(text{Area of } Delta XYZ = frac{24 times 25}{4} = 150 cm^{2})
(therefore text{Area of trapezium PQZY} = 150 – 24 = 126 cm^{2})