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Reproduction in Gymnosperms

Match List I ...

Question

Match List I with List II.

A

A-1, B-2, C-3, D-4

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B

A-1, B-4, C-2, D-3

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C

A-2, B-3, C-4, D-1

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D

A-2, B-3, C-1, D-4

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Solution

The correct option is D
A-2, B-3, C-1, D-4

(D)

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Similar questions

α, β

are the roots of

a x^{2} + b x + c = 0

then match the elements of list I with elements of list II:

List I	List II
A) $\frac{α}{β} + \frac{β}{α} =$	1) $\frac{c^{2}}{a^{2}}$
B) $\frac{α^{2} + β^{2}}{α^{- 2} + β^{- 2}} =$	2) $\frac{c^{5} [3 a b c - b^{3}]}{a^{8}}$
C) $α^{3} + β^{3}$	3) $\frac{b^{2} - 2 a c}{a c}$
D) $α^{5} β^{8} + α^{8} β^{5} =$	4) $\frac{3 a b c - b^{3}}{a^{3}}$

The correct match from List-I to List-II:

A, B, C, D

Q. Suppose four distinct positive numbers

a_{1}, a_{2}, a_{3}, a_{4}

G . P

b_{1} = a_{1}, b_{2} = b_{1} + a_{2}, b_{3} = b_{2} + a_{3}

b_{4} = b_{3} + a_{4}

.
STATEMENT-1 : The numbers

b_{1}, b_{2}, b_{3}, b_{4}

A . P

G . P

.
STATEMENT-2 : The numbers

b_{1}, b_{2}, b_{3}, b_{4}

H . P

.

Q. Assertion :Suppose four distinct positive numbers

a_{1}, a_{2}, a_{3}, a_{4}

b_{1} = a_{1}, b_{2} = b_{1} + a_{2}, b_{3} = b_{2} + a_{3}

b_{4} = b_{3} + a_{4} .

STATEMENT- 1: The numbers

b_{1}, b_{2}, b_{3}, b_{4}

are neither in AP nor in GP, and Reason: STATEMENT- 1: The numbers

b_{1}, b_{2}, b_{3}, b_{4}

Q. Suppose four distinct positive number

a_{1}, a_{2}, a_{3}, a_{4}

are in G.P. Let

b_{1} = a_{1}, b_{2} = b_{1} + a_{2}, a_{3} = b_{2} + a_{3}

b_{4} = b_{3} + a_{4}

Statement 1 - The numbers

b_{1}, b_{2}, b_{3}, b_{4}

are neither in A.P. nor in G.P.
Statement 2 - The numaber

b_{1}, b_{2}, b_{3}, b_{4}

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

(c_{1}, c_{2}, c_{3})

are linearly independent and

x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0

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