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Kaku Goyal an hour ago

Let n = 1
So 1^3 -1 = 0, which is divisible by 6.
Let n = 2
So 2^3 - 2 = 6, which is divisible by 6.
Let n = 3
So 3^3 - 3 = 24, which is divisible by 6.
Hence proved

Posted by Sairubashri T 2 hours ago

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Posted by Amit Jaiswal 3 hours ago

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Gaurav Seth 3 hours ago

given n =11

middle term = (a1 + a11)/2 = 30

a1 + a11 = 60 .......... i

Sn = n/2(a1 + a11) ......... ii

= 11/2 (60) ... from i

= 11 x 30

= 330

Posted by Nitin Kumar 3 hours ago

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Ankit Tarar an hour ago

Firstly take 11 common from this equation than you get == 11(2×3×5×7×1+2). So it is a composite no. Because it has 11 as common and it has three common factor.

Deepak R 3 hours ago

2×3×5×7×11+2
2(1×3×5×7×11+1)
2(15×78)
=2340
Take LCM of 2340
LCM=2×2×5×117

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Soumya Ranjan Mahapatra 4 hours ago

Rationalise denomenator karo answer aa jayegi .....😏😏😏

Posted by Harshdeep Singh 10 hours ago

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Gaurav Seth 10 hours ago

Let 1st term of AP be 'a',

Let the common difference be 'd',

So,as per question, 7 x 7th term = 11 x 11th term

or, 7 [a+6d] = 11 [a+10d]

or, 7a+42d = 11a +110d

or, 4a + 68d = 0

or, a + 17d = 0 .......(1)

Now, 18th term = a+17d

or, t18 = 0 (from 1)

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Harsh Katiyar 4 hours ago

Ye. Less. Then. Ka. Aa. Jaye. Fir. More. Then. Ka. Puch. Lena. Ok

Harsh Katiyar 4 hours ago

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Solve kerke es equation ko badi ajeeb si value aa rhi h 8.95 something😄😄😄😂😂😁😁😀😀

Posted by Aashu Singh 11 hours ago

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Gaurav Seth 11 hours ago

<header data-role="header" data-sticky="true" id="header">

<a> Type your question</a>

</header>Pratik Sabre asked in Math

PQ is a chord of length 8cm to a circle of radius 5cm. The tangents at P and Q intersect at a point T. Find the length TP.

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Rayya Halim answered this

<big id="4134851_vote_span">1732 helpful votes in Math, Class XII-Science</big>

Given, PQ is the chord of the circle and PT and QT are the tangents drawn at the end points of the chord PQ. PQ = 8cm and OP = 5cm.

OT ⊥ PQ,

∴ PR = RQ = (Perpendicular from the centre of the circle to a chord bisect the chord)

In right ΔOPR,

OP2 = PR2 + OR2

⇒ OR2 = OP2 – PR2

⇒ OR2 = (5cm)2 – (4cm2) = (25 – 16)cm2

⇒ OR2 = 9cm2

⇒ OR = 3cm

In right ΔPTR,

PT2 = TR2 + PR2 ...(1)

We know that, the tangent to a circle is perpendicular to the radius through the point of contact.

∴ ∠OPT = 90º

In right ΔOPT,

OT2 = PT2 + OP2 ...(2)

From (1) and (2), we get

OT2 = (TR2 + PR2) + OP2

⇒ (TR + OR)2 = (TR2 + PR2) + OP2

⇒ TR2 + OR2 + 2 × TR × OR = TR2 + PR2 + OP2

⇒ 9 + 6 TR = 16 + 25

⇒ 6TR = 25 + 16 – 9 = 32

⇒ TR =

∴ PT2 = TR2 + PR2 (Using (1))

⇒ PT2 = + (4cm)2

⇒ PT2 =

⇒ PT

Thus, the length of tangent PT is

Posted by Nishu Pradhan 11 hours ago

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Posted by Harsh Gupta 12 hours ago

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Posted by Ritik Kumar 13 hours ago

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Ram Kushwah 12 hours ago

The area of triangle ABC=0

1/2(x(0-b)+a(b-y) +0(y-0))=0

0r -bx+ ab-ay +0=0

bx +by=ab

both side divide by ab

bx/ab + by/ab = ab/ab

x/a +y/b=1

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Posted by Aayushi Tyagi 14 hours ago

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Aditya Bhardwaj 14 hours ago

No because if we take any natural no. Randomly as the power of 4 then it should always be end with 6 or 4

Posted by Ansh Giri 14 hours ago

- 6 answers

Nitin Kumar 5 hours ago

You can see in the cbse.nic .in . Ki site pr maths ka marking scheme ok Ansh

Chetna Pandey 13 hours ago

Check out this app for sample paper woth solution in mathematics section.

Ansh Giri 14 hours ago

Kya tum uske sample paper 1 ke question 6 section A ka solution bata sakte ho plz.

Posted by Devyansh Joshi 14 hours ago

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Kaku Goyal 13 hours ago

(a+b+c)^2=1^2; = a^2+ b^2+ c^2+ 2ab+ 2bc+ 2ca= 1; = ab + bc + ca = 1- a^2- b^2 -c^2 /2; = 1/3
a^2 + b^2 + c^2 = -2/3 + 1 = 1/3
ab + bc + ca = a^2 + b^2 + c^2
On comparison,
a^2 = ab, b^2 = bc, c^2 = ac
So a = b, b = c, c = a
a : b : c = 1 : 1 : 1

Posted by Ansh Giri 14 hours ago

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Ansh Giri 14 hours ago

Code language mein answer bheja hai kya ? Plz mujhe is question ka solution chaiye.

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Saurabh Bisht 14 minutes ago

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