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Algebra
27 Dec 2007 00:44:53 IST
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27 Dec 2007 01:29:23 IST
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Solution(3):
S = [n3 + (n-1)3 + (n-2)3 + ......... + 13] .................
- 2 [(n-1)3 + (n-3)3 + ............... + 23]
= n2(n+1)2/4 - 2*8[(n-1/2)3 + (n-3/2)3 + ........ 13]
= n2(n+1)2/4 - 16*[(n-1/2)3 x {(n-1)/2 + 1}3 / 4
=n2(n+1)2/4 - 4(n-1/2)3.(n-1)/2 + 1}3
On solving we get it as:
= (1/4)(n+1)2(2n-1)
Hope it is useful.
Cheers!!!!!!!!!!!!!!!!!!
S = [n3 + (n-1)3 + (n-2)3 + ......... + 13] .................
- 2 [(n-1)3 + (n-3)3 + ............... + 23]
= n2(n+1)2/4 - 2*8[(n-1/2)3 + (n-3/2)3 + ........ 13]
= n2(n+1)2/4 - 16*[(n-1/2)3 x {(n-1)/2 + 1}3 / 4
=n2(n+1)2/4 - 4(n-1/2)3.(n-1)/2 + 1}3
On solving we get it as:
= (1/4)(n+1)2(2n-1)
Hope it is useful.
Cheers!!!!!!!!!!!!!!!!!!
27 Dec 2007 01:43:00 IST
Like
1 people liked this
Solution(2):
Let side opposite to angle P be "p"
Let side opposite to angle Q be "q"
Let side opposite to angle R Be "r"
Given to us:
SinP , SinQ , SinR are in A.P
Therefore,
p,q,r will also be in A.p as in a triangle the lenght of the
sides are corresponding to their angles.
Let,
SinP/p = SinQ/q = SinR/r = K .......(1)
Also,we know altitudes are
qrSinP/p , prSinQ/q , pqSinR/r
Now using equation (1) we have:
q/pr.SinQ - p/qr.SinP = q/K.pqr - p/k.pqr = q-p/K.pqr
And Similarly:
r/pq.SinR - q/pr.SinQ = r-q/K.pqr
Now we will satisfy condition of p,q,r to be in AP.
So we put
q-p = r-q
we get p,q,r in A.P
Therefore,
r/pq.SinR - q/pr.SinQ = q/pr.SinQ - p/qr.SinP
Now we get,
2(q/pr.SinQ) = r/pq.SinR + p/qr.SinP
So we have,
p/qr.SinP , q/pr.SinQ , r/pq.SinR in A.P
And
qr.SinP/p , pr.SinQ/q , pq.SinR/r in H.P
Hence answer...
altitudes are in H.P
Proved.
Hope you find it useful.
Cheers!!!!!!!!!@@@!!!!!!!!!!
Let side opposite to angle P be "p"
Let side opposite to angle Q be "q"
Let side opposite to angle R Be "r"
Given to us:
SinP , SinQ , SinR are in A.P
Therefore,
p,q,r will also be in A.p as in a triangle the lenght of the
sides are corresponding to their angles.
Let,
SinP/p = SinQ/q = SinR/r = K .......(1)
Also,we know altitudes are
qrSinP/p , prSinQ/q , pqSinR/r
Now using equation (1) we have:
q/pr.SinQ - p/qr.SinP = q/K.pqr - p/k.pqr = q-p/K.pqr
And Similarly:
r/pq.SinR - q/pr.SinQ = r-q/K.pqr
Now we will satisfy condition of p,q,r to be in AP.
So we put
q-p = r-q
we get p,q,r in A.P
Therefore,
r/pq.SinR - q/pr.SinQ = q/pr.SinQ - p/qr.SinP
Now we get,
2(q/pr.SinQ) = r/pq.SinR + p/qr.SinP
So we have,
p/qr.SinP , q/pr.SinQ , r/pq.SinR in A.P
And
qr.SinP/p , pr.SinQ/q , pq.SinR/r in H.P
Hence answer...
altitudes are in H.P
Proved.
Hope you find it useful.
Cheers!!!!!!!!!@@@!!!!!!!!!!
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Solution(1):
Cos(x-y) , Cosx , Cos(x+y) are in HP
So we have :
Cosx = 2 Cos(x-y) Cos(x+y) / Cos(x-y) + Cos(x+y)
Cosx = 2(Cos2x - Sin2y) / 2 Cosx.Cosy
In the above step expanding and solving with Formula.
Cos(A+B) and Cos(A-B)
Cosx(Cosx.Cosy) = (Cos2x - Sin2y)
Cos2x.Cosy = (Cos2x - Sin2y)
Cos2x.Cosy - Cos2x = (-Sin2y)
Cos2x (Cosy - 1) = (-Sin2y)
OR
Cos2x(1-Cosy) = Sin2y
Now we know that,
Sin2A + Cos2A = 1
So,
Sin2y + Cos2y = 1
Sin2y = 1 - Cos2y
So we have,
Cos2x(1-Cosy) = 1 - Cos2y
Cos2x(1-Cosy) = (1)2-Cos2y
In abv step applying for A2-B2 = (A+B)(A-B)
Cos2x (1-Cosy) = (1+Cosy)(1-Cosy)
Cos2x = 1 + Cosy
Now we know that,
Cos2A = 2Cos2A - 1
Cos2A + 1 = 2Cos2A
so we get:
1+Cosy = 2Cos2(y/2)
Cos2x = 2Cos2(y/2)
Cos2x.[ 1/Cos2(y/2) ] = 2
Cos2x.Sec2(y/2) = 2
Taking Square root,
Cosx.Sec(y/2) =
Hope You find it useful.
Cheers!!!!!!!!!!!!!!