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Ask QuestionPosted by Roshni Soren 6 years, 5 months ago
- 2 answers
Gaurav Seth 6 years, 5 months ago
Step-by-step explanation: 6,12,18.......
a=6
d=6
n=10
s=10/2[2*6+(10-1)(6)]
s=5[12+54]
s=5[66]
s=330
Posted by Sireesha Reddy 6 years, 5 months ago
- 1 answers
Gaurav Seth 6 years, 5 months ago
The Given Series : x+b, x+3b, x + 5b,...
This is an A.P.
∴ First Term, a = x+b
Difference, d = x+3b - (x+b) = x+3b - x-b= 2b
Now, The General Term ,
= a+(n-1)d
= x+b+(n-1)2b
= x+b+(2n-2)b
= x+b(1+2n-2)
= x+b(2n-1)
a = x+b
Posted by Maharnab Goswami 6 years, 5 months ago
- 1 answers
Gaurav Seth 6 years, 5 months ago
l = 28
S = 144
n = 9
We have to find a,
The value of a = 4
Posted by Sahil Kumar 6 years, 5 months ago
- 0 answers
Posted by Tiger Saraf 6 years, 5 months ago
- 1 answers
Yogita Ingle 6 years, 5 months ago
Let us assume that √5 is a rational number.
we know that the rational numbers are in the form of p/q form where p,q are integers.
so, √5 = p/q
p = √5q
we know that 'p' is a rational number. so √5 q must be rational since it equals to p
but it doesnt occurs with √5 since its not an integer
therefore, p =/= √5q
this contradicts the fact that √5 is an irrational number
hence our assumption is wrong and √5 is an irrational number.
Posted by Thahira Begum 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Check last year papers here : https://mycbseguide.com/cbse-question-papers.html
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Mr. Spoidormon 6 years, 5 months ago
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Alan D Sam 6 years, 5 months ago
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Alan D Sam 6 years, 5 months ago
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Posted by Md Aarif 6 years, 5 months ago
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Yogita Ingle 6 years, 5 months ago
We applied Euclid Division algorithm on n and 3.
a = bq +r on putting a = n and b = 3
n = 3q +r , 0<r<3
i.e n = 3q -------- (1),n = 3q +1 --------- (2), n = 3q +2 -----------(3)
n = 3q is divisible by 3
or n +2 = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Posted by Rishita Golchha 6 years, 5 months ago
- 0 answers
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Jaimina Gharia 6 years, 5 months ago
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Ayesha Jawed 6 years, 5 months ago
Posted by Parag Deshpande 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
{tex}\frac{105}{10}=10.5{/tex}
Hence 105 is not divisible by 10
But LCM of two numbers should be divisible by their HCF.
{tex}\therefore{/tex} Two numbers cannot have their HCF as 10 and LCM as 105.
Posted by Parag Deshpande 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Using the result,
{tex}HCF{\text{ }} \times {\text{ }}LCM{/tex} = Product of two natural numbers
{tex} \Rightarrow {/tex} the other number = {tex}\frac{{9 \times 90}}{{18}}{/tex} = 45
Posted by Chethan Chethan 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
The given n odd natural numbers are,
1, 3, 5, .... n
Clearly,it is an AP with ,
First term (a) = 1
Common difference (d) = 3 - 1 = 2
Number of terms = {tex}{/tex}n
Sum of n terms {tex} = \frac{n}{2}\left[ {2a + (n - 1)d} \right]{/tex}
{tex} = \frac{n}{2}\left[ {2 \times 1 + (n - 1) \times 2} \right]{/tex}
{tex} = \frac{n}{2}\left[ {2 + 2n - 2} \right]{/tex}
{tex} = \frac{n}{2} \times 2n{/tex}
= n2
Posted by Khushboo Kataria 6 years, 5 months ago
- 1 answers
Rajan Kumar Pasi 6 years, 5 months ago
Please learn the trigonometric formulae, this is a very simple question to ask .
{tex}\large Question: \space To\space prove:cos^2A +\frac{1}{1+cot^2A} =1?{/tex}
{tex}\large{ Solution\implies LHS \implies cos^2A +\frac1{1+cot^2A} }{/tex}
{tex}\large{ \big(1+cot^2A = cosec^2A \big) }{/tex}
{tex}\large{ \implies cos^2A +\frac1{\underline {1+cot^2A}}\implies cos^2A+\frac 1 { cosec^2A} }{/tex}
{tex}\large{ \big(\frac1{cosec\space A} = sin\space A \big)\\ \big(\frac1{cosec^2A} = sin^2A \big) }{/tex}
{tex}\large{ \implies cos^2A+sin^2A }{/tex}
{tex}\large{ \big(cos^2A+sin^2A=1 \big) }{/tex}
{tex}\large{Ans \implies \boxed1 }{/tex}
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Sumit Jangra 6 years, 5 months ago
0Thank You