Find the sum of all 3 digit number which are multiple of 7
Answers
Answered by
120
Smallest three digit number divisible by 7 = 105 and the largest = 994
An = 994
A = 105
D = 7
n= ?
An = A + [ n - 1] D
thus
994 = 105 + [ n - 1 ] 7
thus by solving further
n = 128
now,
Sn = n/2 [ A + An]
Sn = 128/2 [ 105 + 994]
thus,
Sn = 70336.
Thus the sum of all the three digit numbers divisible by 7 is 70336.
An = 994
A = 105
D = 7
n= ?
An = A + [ n - 1] D
thus
994 = 105 + [ n - 1 ] 7
thus by solving further
n = 128
now,
Sn = n/2 [ A + An]
Sn = 128/2 [ 105 + 994]
thus,
Sn = 70336.
Thus the sum of all the three digit numbers divisible by 7 is 70336.
Answered by
36
First three digit term divisible by 7 is 105
Last three digit term divisible by 7 is 994
We know that tn=a+(n−1)dtn=a+(n−1)d
⇒994=105+(n−1)7⇒994=105+(n−1)7
⇒n=128⇒n=128
We know that Sn=n2Sn=n2(l+a)(l+a)
∴∴ The sum of the required series Sn=1282(994+105)Sn=1282(994+105)
=64×1099=70336
Last three digit term divisible by 7 is 994
We know that tn=a+(n−1)dtn=a+(n−1)d
⇒994=105+(n−1)7⇒994=105+(n−1)7
⇒n=128⇒n=128
We know that Sn=n2Sn=n2(l+a)(l+a)
∴∴ The sum of the required series Sn=1282(994+105)Sn=1282(994+105)
=64×1099=70336
Similar questions