Match the following by appropriately matching the lists based on the information given in Column I and Column II -array - - Wline - bff Column - I - Wine- oga 10- logb 10- log - c 10 text are in 1 WlineWine-a- x- b text are in G-P- and -b- y - c text are in G-P- 11Wline-endarray -A- a - q - b - p - c - rB- a - p - b - r - c - qC- a - q - b - r - c - pD - a - r - b - q - c - p

The correct option is C a → q, b nbsp;→ r, c nbsp;→ pa. a, b, c are in G.P. b^2 =ac ⇒ 2 log_10 b = log_10 a +log_10 c ⇒2log_b 10= nbsp;1log_a 10+1log_c 1 ...

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Byju's Answer

Standard XII

Mathematics

Progression

Match the fol...

Question

Match the following by appropriately matching the lists based on the information given in Column I and Column II
$\begin{matrix} C o l u m n I & C o l u m n I I a . & If a, b, c are in G . P ., then & p. & A . P . {log}_{a} 10, {log}_{b} 10, {log}_{c} 10 are in b . & If \frac{a + b e^{x}}{a - b e^{x}} = \frac{b + c e^{x}}{b - c e^{x}} = \frac{c + d e^{x}}{c - d e^{x}}, & q. & H . P . then a, b, c, d c . & If a, b, c are in A . P .; & r. & G . P . a, x, b are in G . P . and b, y, c are in G . P ., then x^{2}, b^{2}, y^{2} are in \end{matrix}$

a \to p, b \to r, c \to q

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a \to q, b \to p, c \to r

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a \to q, b \to r, c \to p

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a \to r, b \to q, c \to p

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Solution

The correct option is C $a \to q, b \to r, c \to p$
$a .$
$a, b, c$ are in G.P.
$b^{2} = a c \Rightarrow 2 {log}_{10} b = {log}_{10} a + {log}_{10} c \Rightarrow \frac{2}{{log}_{b} 10} = \frac{1}{{log}_{a} 10} + \frac{1}{{log}_{c} 10}$
So,
${log}_{a} 10, {log}_{b} 10, {log}_{c} 10$ are in H.P.
$(a) \to (q)$

$b .$
$\frac{a + b e^{x}}{a - b e^{x}} = \frac{b + c e^{x}}{b - c e^{x}} = \frac{c + d e^{x}}{c - d e^{x}} \Rightarrow \frac{a + b e^{x}}{a - b e^{x}} + 1 = \frac{b + c e^{x}}{b - c e^{x}} + 1 = \frac{c + d e^{x}}{c - d e^{x}} + 1 \Rightarrow \frac{2 a}{a - b e^{x}} = \frac{2 b}{b - c e^{x}} = \frac{2 c}{c - d e^{x}} \Rightarrow \frac{a - b e^{x}}{a} = \frac{b - c e^{x}}{b} = \frac{c - d e^{x}}{c} \Rightarrow 1 - \frac{b}{a} e^{x} = 1 - \frac{c}{b} e^{x} = 1 - \frac{d}{c} e^{x} \Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$
Hence $a, b, c, d$ are in G.P.
$(b) \to (r)$

$c .$
$2 b = a + c x^{2} = a b y^{2} = b c \Rightarrow x^{2} + y^{2} = b (a + c) = 2 b^{2} \Rightarrow b^{2} = \frac{x^{2} + y^{2}}{2}$
Hence $x^{2}, b^{2}, y^{2}$ are in A.P.
$(c) \to (p)$

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