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Yogita Ingle 6 years, 2 months ago
centeroid of a triangle is x=( x1+x2+x3÷3); y= (y1+y2+y3÷3)
so here [-5+4+(-k)]÷3 = k-1
-1-k = 3(k-1)
-1-k = 3k -3
-1 + 3 = 3k+k
2=4k
k= 1/4 = 1/2
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Yogita Ingle 6 years, 2 months ago
Let N be a number such that,
N = 5( mod 16)
i.e.
N = 16x + 5
Where, x is any positive integer.
We can write,
N = 8(2x) + 5
if x = positive integer, 2x is also a positive integer.
Let 2x = y,
N = 8y + 5
Thus, N = 5(mod 8)
Hence, the remainder when the same number is divided by 8 would also be 5.
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Yogita Ingle 6 years, 2 months ago
Let suppose the point on Y axis is A
So the A(0,y)
(5,-2) = P
(-3,2)=Q
AP²= AQ²..(Equidistant point)
By distance formula
(0-5)² + [(y-(-2)]²=[(0-(-3)]² + (y-2)²
25+y²+4y+4 = 9+ y²-4y+4
4y +4y=9-25 ......(like term)
8y=-16
y=-16/8
y=-2
A(0,-2)
Posted by Abhinav Thapliyal 6 years, 2 months ago
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Posted by Harsh Kumar 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
We can say that 30 is the Median of 11 terms.
therefore
we know
M = (n + 1)/2 th observation for n = odd
Therefore the 6th term of this AP is 30
Therefore A6 = 30
a + 5d = 30
therefore we need to find S11
Therefore S11 = n/2(2a + (n-1)d)
=11/2(2a + 10d)
=11/2(2(a + 5d)
replacing value
=11/2(2(30))
=11 * 30
=330
therefore Answer is 330
Posted by Sakshi Vats 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
First term, a = 24
common difference, d = 21-24 = -3
let the number of terms to get sum 78 is n.
{tex} S_n=78\\ \\ \Rightarrow \frac{n}{2}(2a+(n-1)d)=78\\ \\ \Rightarrow \frac{n}{2}(2 \times 24+(n-1)(-3))=78\\ \\ \Rightarrow n(48-3n+3)=78 \times 2\\ \\ \Rightarrow -3n^2+51n-156=0\\ \\ \Rightarrow 3n^2-51n+156=0{/tex}
Solving the quadratic equation, we get n=4 and n=13.
So you can take either 4 terms or 13 terms to get the sum 78.
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