What is the average of all numbers between 9 and 90 which are divisible by 8?
Answers
Answered by
1
The number between 9 and 90 which are divisible by 8 are:
16,24,32,40,48,56,64,72,80,88
They are 10 in number
We get their average by adding them together then dividing by 10
Average = (16+24+32+40+48+56+64+72+80+88)/10
= 520/10
= 52
16,24,32,40,48,56,64,72,80,88
They are 10 in number
We get their average by adding them together then dividing by 10
Average = (16+24+32+40+48+56+64+72+80+88)/10
= 520/10
= 52
Answered by
2
Hey! ! !
Solution :-
☆This is a problem of AP.
Here the first term which is divisible by 8 after 10 is 16. So the first term a of the AP is 16. And the common difference d is 8. Last term of the AP is 88.
Last term of an AP is given by the formula
Tn = a+((n-1)*d)
Using that formula, we get
88 = 16 + ((n-1)*8)
88–16 = 72 = (n-1) *8
n-1 = 72/8 = 9
n-1 = 9
n = 9+1 = 10
☆ ☆ ☆ Hop its helpful ☆ ☆ ☆
☆ Regards :- ♡♡《 Nitish kr singh 》♡♡
Solution :-
☆This is a problem of AP.
Here the first term which is divisible by 8 after 10 is 16. So the first term a of the AP is 16. And the common difference d is 8. Last term of the AP is 88.
Last term of an AP is given by the formula
Tn = a+((n-1)*d)
Using that formula, we get
88 = 16 + ((n-1)*8)
88–16 = 72 = (n-1) *8
n-1 = 72/8 = 9
n-1 = 9
n = 9+1 = 10
☆ ☆ ☆ Hop its helpful ☆ ☆ ☆
☆ Regards :- ♡♡《 Nitish kr singh 》♡♡
Anonymous:
: (
Nthig
: )
This solution is wrong
Similar questions