Question

please solve this...........​

Answer 1

Question 1 :-

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Find the first four terms of an A.P. where the first term is -2 and the common difference is also -2.

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Solution :-

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Given :

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$\sf \large{ t_{1} \: = \: - 2} \\ \sf \large{d \: = \: - 2 \: }$

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To find :

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$\sf \large{ t_{2} = \: \: ?} \\ \sf \large{ t_{3} = \: \: ?} \\ \sf \large{ t_{4} = \: \: ?}$

Answer :

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We know that ,

$\sf \large \red{ d \: = \: t_{2} \: - \: t_{1} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \sf \large{\therefore \: t_{2} = d \: + \: t_{1} = - 2 - 2 = \purple{- 4}}$

Similarly,

$\sf \large \red{ d \: = \: t_{3} \: - \: t_{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \sf \large{\therefore \: t_{3} = d \: + \: t_{2} = - 2 -4 = \purple{- 6}}$

$\sf \large \red{ d \: = \: t_{4} \: - \: t_{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \sf \large{\therefore \: t_{4} = d \: + \: t_{3} = - 2 -6 = \purple{- 8}}$

Hence,

The correct option is Option C

-2 , -4 , -6 , -8 are the first four terms of the A.P.

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Question 2 :-

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The side of a square is 12 cm . Find the length of its diagonal.

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Solution :-

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We know that,

$\bf \large{Side \: of \: a \: square = \dfrac{Diagonal}{ \sqrt{2} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \bf \large{12 = \dfrac{Diagonal}{ \sqrt{2}}} \\ \\ \: \: \: \: \: \: \: \implies \bf \large{12 \sqrt{2} = Diagonal}\\ \\ \bf \large{ \therefore \: Diagonal \: of \: the \: square \: is \: \red{12 \sqrt{2} \: cm }}$

Hence, Option B is the correct option

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Question 3 :-

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If a pair of linear equations is consistent, then the lines will be

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Solution :-

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If a pair of linear equations is consistent, then the lines will be intersecting or coincident

Hence, Option D is the correct option.

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Question 4 :-

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PQ and PR are drawn to a circle as shown in the figure and ∠QPR = 90° and PQ = 8cm then find the radius of the circle.

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Solution :-

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$\bf\Large\red{Refer \: to \: the \: 1st \: attachment \: !}$

The correct option is Option A

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Question 5 :-

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Write the co-ordinates of the mid point of the line joining (4,10) and (-2,4)

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Solution :-

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$\bf\Large\purple{Refer \: to \: the \: 2nd \: attachment \: !}$

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Answer 2

Answer:

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