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Match the following.
LIST-I
LIST-II
A)
The sequence $a, b, 10, c, d$ is an arithmetic progression. The value of $a + b + c + d$
P)
10
B)
The sides of right triangle form a three term geometric sequence. The shortest side has length 2. The length of the hypotenuse is of the form $a + \sqrt{b}$ where $a \in N$ and $\sqrt{b}$ is a surd, then value of $a^{2} + b^{2}$ is
Q)
20
C)
The sum of first three consecutive numbers of a decreasing infinite G.P. is 70, if the two extremes be multiplied each by 4, and the mean by 5, the resultant numbers are in A.P. The first term of the G.P. is
R)
26
D)
The diagonals of parallelogram have a measure of 4 and 6 metres. They cut off forming an angle of $60^{0}$ . If the perimeter of the parallelogram $2 (\sqrt{a} + \sqrt{b})$ is where $a, b \in N$ then $(a + b)$ equals
S)
40

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$A - S; B - R; C - S; D - R$

$A - R; B - S; C - S; D - R$

$A - S; B - R; C - R; D - S$

$A - R; B - R; C - S; D - S$

answer is A.

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Detailed Solution

A: b + c = 20 and a + d = 20
$\Rightarrow a + b + c + d = 40$
B : Sides are $2, 2 r, 2 r^{2}$
Now ${(2 r)}^{2} = 4 r^{2} + 4$
$\Rightarrow r^{4} - r^{2} - 1 = 0$
$\Rightarrow r^{2} = \frac{1 + \sqrt{5}}{2}$
$\Rightarrow 2 r^{2} = 1 + \sqrt{5}$
$a^{2} + b^{2} = 26$
C: $a + a r + a r^{2} = 70$
Now, 4a, 5ar, $4 a r^{2}$ are in A. P.
$\Rightarrow 5 r - 4 = 4 r^{2} - 5 r \Rightarrow C_{1}^{2} = 7$
$\Rightarrow 5 r = 2 (r^{2} + 1)$ $\Rightarrow C_{1} = \sqrt{7}$
So, $a r + a (\frac{5 r}{2}) = 70$ Other side is $\cos 120^{0} = \frac{4 + 9 - C_{2}^{2}}{12}$
$\Rightarrow a r = 20$ $\Rightarrow C_{2}^{2} = 19$
$\Rightarrow$ G. P. is 40, 20, 10 $\Rightarrow C_{2} = \sqrt{19}$
Perimeter = 2 $(\sqrt{7} + \sqrt{19})$
D : $\cos 60^{0} = \frac{4 + 9 - C_{1}^{2}}{12}$ $\Rightarrow a + b = 26$

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