(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 diag...
Question 1 Report
(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.
Answer Details
(a) Arithmetic progression.
Let the first term be \(a\) and the common difference be \(d\). The \(n\)th term is \(T_n = a + (n-1)d\).
Given \(T_3 = -9\) and \(T_8 = 26\):
\[a + 2d = -9 \quad\text{...(1)}\]\[a + 7d = 26 \quad\text{...(2)}\](i) Common difference. Subtract (1) from (2):
\[5d = 35 \quad\Rightarrow\quad d = 7\](ii) First term. Substitute \(d = 7\) into (1):
\[a + 2(7) = -9 \quad\Rightarrow\quad a + 14 = -9 \quad\Rightarrow\quad a = -23\]Common difference \(= 7\); first term \(= -23\).
(b) Area of the trapezium PQZY.
From the diagram, \(PQ \parallel YZ\), with \(|XP| = 2\text{ cm}\), \(|PY| = 3\text{ cm}\), \(|PQ| = 6\text{ cm}\) and area of \(\triangle XPQ = 24\text{ cm}^2\). P lies on XY and Q lies on XZ.
Since \(PQ \parallel YZ\), triangle \(XPQ\) is similar to triangle \(XYZ\) (equal corresponding angles). The ratio of corresponding sides along XY is:
\[\frac{XP}{XY} = \frac{XP}{XP + PY} = \frac{2}{2 + 3} = \frac{2}{5}\]For similar triangles the ratio of areas is the square of the ratio of sides:
\[\frac{\text{Area }\triangle XPQ}{\text{Area }\triangle XYZ} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}\]Therefore:
\[\text{Area }\triangle XYZ = 24 \times \frac{25}{4} = 150 \text{ cm}^2\]The trapezium PQZY is what remains when \(\triangle XPQ\) is removed from \(\triangle XYZ\):
\[\text{Area of trapezium PQZY} = 150 - 24 = 126 \text{ cm}^2\]Area of trapezium PQZY \(= 126\text{ cm}^2\).
Download The App On Google Playstore
Everything you need to excel in your exams
