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Ask QuestionPosted by Santa Bir Gajmer 5 years, 3 months ago
- 2 answers
Amit Kumar Chaurasia 5 years, 3 months ago
Gaurav Seth 5 years, 3 months ago
| 12576 | <a href="http://www.leastcommonmultiple.net/lcm-of-2/">2</a> |
| 6288 | <a href="http://www.leastcommonmultiple.net/lcm-of-2/">2</a> |
| 3144 | <a href="http://www.leastcommonmultiple.net/lcm-of-2/">2</a> |
| 1572 | <a href="http://www.leastcommonmultiple.net/lcm-of-2/">2</a> |
| 786 | <a href="http://www.leastcommonmultiple.net/lcm-of-2/">2</a> |
| 393 | <a href="http://www.leastcommonmultiple.net/lcm-of-3/">3</a> |
| 131 | 131 |
| 1 |
LCM = 25. 31. 1311 = 12576
Posted by Nilam Dang 5 years, 3 months ago
- 0 answers
Posted by Sumaila Ali Choudhary??? 5 years, 3 months ago
- 1 answers
Gaurav Seth 5 years, 3 months ago
It is given that ,
x ,2x+k and 3x+6 are three consecutive terms .
we know that ,
Difference of any two consecutive terms is equal in A.P
Now,
(2x+k)-x = (3x+6)-(2x+k)
=> 2x+k-x = 3x+6-2x-k
=> x+k = x+6-k
=> k+k = x+6-x
=> 2k = 6
=>k = 6/2
=> k = 3
Therefore,
If k = 3 then the given three consecutive terms x,2x+k and 3x+6 are in A.P
Posted by Mehul Patel 5 years, 3 months ago
- 2 answers
Posted by Aditya Shejal 5 years, 3 months ago
- 1 answers
Gaurav Seth 5 years, 3 months ago

Given a ∆ABC such that D is the mid point of AB and DE || BC.
To Prove: AE = EC
Proof: AD = DB (given)

In ∆ABC, we have
DE || BC
Therefore, by using Basic proportionality theorem, we have

Comparing (i) and (ii), we get


Posted by Priyanshu Pandit 5 years, 3 months ago
- 0 answers
Posted by Sourav Keshri 5 years, 3 months ago
- 5 answers
Gaurav Seth 5 years, 3 months ago
Let √2 be a rational number
Therefore, √2= p/q [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0
On squaring both sides, we get
p²= 2q² ...(1)
Clearly, 2 is a factor of 2q²
⇒ 2 is a factor of p² [since, 2q²=p²]
⇒ 2 is a factor of p
Let p =2 m for all m ( where m is a positive integer)
Squaring both sides, we get
p²= 4 m² ...(2)
From (1) and (2), we get
2q² = 4m² ⇒ q²= 2m²
Clearly, 2 is a factor of 2m²
⇒ 2 is a factor of q² [since, q² = 2m²]
⇒ 2 is a factor of q
Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1
Therefore, Our supposition is wrong
Hence √2 is not a rational number i.e., irrational number.
Posted by Kanika Mehra Grover 5 years, 3 months ago
- 2 answers
Posted by Shivani Suraj 5 years, 3 months ago
- 5 answers
Posted by Akash Singh Rajput 5 years, 3 months ago
- 4 answers
Posted by Vanshika Baliyan 5 years, 3 months ago
- 4 answers
Posted by Suryansh Singh 5 years, 3 months ago
- 2 answers
Posted by Nayasa . 5 years, 3 months ago
- 1 answers
Posted by Umang Katàçgc 5 years, 3 months ago
- 0 answers
Posted by Amit Tiger 5 years, 3 months ago
- 1 answers
Posted by Amit Tiger 5 years, 3 months ago
- 0 answers
Posted by Karnika Singh 5 years, 3 months ago
- 3 answers
Posted by Bishant Pandey 5 years, 3 months ago
- 3 answers
Jags Sma 5 years, 3 months ago
Yogita Ingle 5 years, 3 months ago
(a + b)2 = (a + b) (a + b)
= a(a+b) + b(a +b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
Posted by Vanshika Baliyan 5 years, 3 months ago
- 1 answers
Tarannum Jahan 5 years, 3 months ago
Posted by Aayan Gupta 5 years, 3 months ago
- 2 answers
Jags Sma 5 years, 3 months ago
Posted by Suraj Kumar 5 years, 3 months ago
- 0 answers
Posted by Sunny Shokeen 5 years, 3 months ago
- 5 answers
Posted by Nitesh Deswal 5 years, 3 months ago
- 5 answers
Bhawna Koranga 5 years, 3 months ago
Vidya Sagar Bonu 5 years, 3 months ago
Posted by Teenu Katoch 5 years, 3 months ago
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Posted by Yash Khatode 5 years, 3 months ago
- 2 answers
Jags Sma 5 years, 3 months ago
Gaurav Seth 5 years, 3 months ago
Show that n2 - 1 is divisible by 8, if n is an odd positive integer.
Any odd positive integer n can be written in form of 4q + 1 or 4q + 3.
If n = 4q + 1, when n2 - 1 = (4q + 1)2 - 1 = 16q2 + 8q + 1 - 1 = 8q(2q + 1) which is divisible by 8.
If n = 4q + 3, when n2 - 1 = (4q + 3)2 - 1 = 16q2 + 24q + 9 - 1 = 8(2q2 + 3q + 1) which is divisible by 8.
Posted by Misba K M 5 years, 3 months ago
- 0 answers
Posted by Disha Rohage 5 years, 3 months ago
- 2 answers
Bhawna Koranga 5 years, 3 months ago

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