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Gaurav Seth 6 years, 6 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.
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Sia ? 6 years, 6 months ago
The general term of an AP is given by an=a+(n-1)d and Sn={tex}\frac{n}{2}{/tex}[2a+(n-1)d].
Given that a2=14 and a3=18
So, d=a3-a2=18-14=4
Now, a2=14 {tex}\Rightarrow{/tex}a+4=14 {tex}\Rightarrow{/tex}a=10
Also, S51={tex}\frac{{51}}{2}{/tex}[2(10)+(50)4]
{tex}\Rightarrow{/tex}S51={tex}\frac{{51}}{2}{/tex}[20+200]
{tex}\Rightarrow{/tex}S51={tex}\frac{{51}}{2}{/tex}[220]
{tex}\Rightarrow{/tex}S51=51{tex}\times{/tex}110
{tex}\Rightarrow{/tex}S51=5610
1Thank You