⇔⇔⇔→Only for special ranked← ⇔⇔⇔
If x is the average (arithmetic mean) of m and 9, y is the average of 2m and 15, and z is the average of 3m and 18, what is the average of x, y, and z in terms of m?
↓
→Content quality required← This one is for 100 points , i need a very good answers
↑
Thanks :)
Answers
Answered by
2
Here is your required answer
By Adeeb Khan
3/2 (m+7)
See the steps I attachment
Attachments:
Hakar:
thanks for helped me sir/ma'am :)
It's my pleasure.
Answered by
1
Solution :
****************************************
We know that ,
If p , q , r in A.P , then mean of
p and r is q.
q = ( p + r )/2
*******************************************
According to the problem given ,
i ) mean of m and 9 = x
x = ( m + 9 )/2 ----( 1 )
ii ) mean of 2m and 15 = y
y = ( 2m + 15 )/2 ----( 2 )
iii ) mean of 3m and 18 = z
=> z = ( 3m + 18 )/2 ---( 3 )
iv ) mean of x , y and z = ( x+y+z)/3
[ from ( 1 ) ,(2 ) and ( 3 ) ]
= [(m+9)/2 + (2m+15)/2 + (3m+18)/2]/3
= [ m+9+2m+15+3m+18 ]/( 2 × 3 )
= ( 6m + 42 )/6
= [ 6( m + 7 )/6 ]
= m + 7
••••
****************************************
We know that ,
If p , q , r in A.P , then mean of
p and r is q.
q = ( p + r )/2
*******************************************
According to the problem given ,
i ) mean of m and 9 = x
x = ( m + 9 )/2 ----( 1 )
ii ) mean of 2m and 15 = y
y = ( 2m + 15 )/2 ----( 2 )
iii ) mean of 3m and 18 = z
=> z = ( 3m + 18 )/2 ---( 3 )
iv ) mean of x , y and z = ( x+y+z)/3
[ from ( 1 ) ,(2 ) and ( 3 ) ]
= [(m+9)/2 + (2m+15)/2 + (3m+18)/2]/3
= [ m+9+2m+15+3m+18 ]/( 2 × 3 )
= ( 6m + 42 )/6
= [ 6( m + 7 )/6 ]
= m + 7
••••
: )
thanks for helped me sir/ma'am :)
Similar questions