x+y=15 then x+y+z=z+15
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MATHS
The value of x+y+z=15, if a,x,y,z,b are in A.P, while the value of
x
1
+
y
1
+
z
1
=
3
5
if a,x,y,z,b are in H.P. Find a× b.
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ANSWER
To find the value of a and b
Given,
x+y+z=15....(i)
x
1
+
y
1
+
z
1
=
3
5
........(ii)
if a,x,y,z are in A.P
Then, x,y,z are in A.P
∴2y=x+z......(iii)
⇒2y+y=15........... comparing (iii) and (i)
⇒3y=15⇒
y=5
Now,
2
a+b
=y⇒a+b=2y
⇒a+b=10......(vi)
Now, if a,x,y,z are in HP
then x,y,z one in H.P
∴
y
2
=
x
1
+
z
1
.......(iv)
⇒
x
1
+
y
1
+
z
1
=
3
5
.......(v)
from (iv) and (v) we can write
y
2
+
y
1
=
3
5
=
3
5
........putting (iv) in (v)
y
3
=
3
5
⇒
y=
5
9
Also, a,y,b are in H.P.
a
1
+
b
1
=
y
2
⇒
ab
a+b
=
y
−2
=
9
2×5
⇒
ab
a+b
=
9
10
......(vi)
by comparing (vi) and (vii) we can say that
ab
a+b
=
9
10
⇒
ab
10
=
9
10
⇒
ab=9
.......(viii)
⇒a(10−a)=9
⇒10a−a
2
=9⇒a
2
−10a+9=0
⇒(a−1)(a−9)=0
∴a=1 or 9
∴ when a=1,b=9
a=9,b=1
∴
a×b=9×1=1×9=9
✓ Given
x + y = 15
✓ To find
x + y + z = z + 15
✓ Solution
x+y = 15 ----- [ 1 ]
(x + y) + z = z + 15 -----[ 2 ]
Let us substitute equation [1] in [2]
15 + z = z + 15
Here even the variable gets cancelled !
So , if z takes place here then it takes 0 as its value.
z = 0
[ or ]
No z exists { since it is been eliminated }