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Posted by Guruveer Singh 5 years, 5 months ago
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Yogita Ingle 5 years, 5 months ago
The first 8 multiples of 3 are : 3, 6, 9, 12, 15, 18, 21, 24
Number of terms = 8
The first term = a1 = 3
The last term = a8 = 24
Find the sum:
Sn = n/2 ( a1 + an)
S8 = 8/2 (3 + 24)
S8 = 108
Hence the sum of the first 8 multiples of 3 is 108.
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Lucifer ? Morningstar? 5 years, 5 months ago
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Yuvika Mourya 5 years, 5 months ago
Lucifer ? Morningstar? 5 years, 5 months ago
Posted by Lucifer?? Morningstar?? 5 years, 5 months ago
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Yogita Ingle 5 years, 5 months ago
Let a , d and A , D be the first term and the common ratio of the first and the second AP respectively.
So [(n/2){2a+(n-1)d}]/[(n/2){2A+(n-1)D}]
=(7n-5)/(5n+17)={7(n-1)+2}/{5(n-1)+22}
So {2a+(n-1)d}/{2A+(n-1)D}
={2+7(n-1)}/{22+5(n-1)} =>
a=1, d=7, A =11, D = 5
6th term of 1st AP =a+5d=1+5×7=36
6th term of 2nd AP=A+5D=11+5×5=36
Posted by Ankit Kumar 5 years, 5 months ago
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Posted by Debraj Chanda 5 years, 5 months ago
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Yogita Ingle 5 years, 5 months ago
2x+3y=7
ax−bx+ay+by=3a+b−2
(or) 2x+3y=7 …………(1)
(a−b)x+(a+b)y=3a+b−2 ………….(2)
Given: the pair of linear equations have infinite number of solutions
2/ a−b = 3/ a−b = 7/ 3a+b−2
⇒2/ a−b = 3/ a+b
3/ a+b =7/ 3a+b−2
⇒2a+2b=3a−3b
⇒−a=−5b
∴a=5b
3(3a+b−2)=7(a+b)
9a+3b−6=7a+7b
2a−4b−6=0
a−2b=3
Put a=5b in a−2b=3 we get
5b−2b=3
⇒3b=3
or b=1
∴a=5×1=5, b=1
Hence a=5,b=1.
Posted by Vinit Chaudhary 5 years, 5 months ago
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Posted by Khushboo Tiwari 5 years, 5 months ago
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Khushboo Tiwari 5 years, 5 months ago
Posted by Khushboo Tiwari 5 years, 5 months ago
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Yogita Ingle 5 years, 5 months ago
Distance between = √(x2-x1)2 + (y2-y1)2
=√(0-cos2θ ) +(sin2θ -0)
=√-cos2θ+sin2θ
= √sin2θ -cos2θ
Posted by Khushboo Tiwari 5 years, 5 months ago
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Yogita Ingle 5 years, 5 months ago
factors of 120
120 = 2×2×2×3×5×1
factors of 121
121= 11×11×1
multiply all the common prime factors by the lowest exponents,
but these two numbers have no common prime factors.
common factors of 120 and 121 is only 1
therefore ,
HCF of 120 and 121 is 1
hence,
HCF of 120 and 121 = 1
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