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D
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Difficulty:
35%
(medium)
Question Stats:
87% (01:36) correct
13%
(01:38)
wrong
based on 38
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What is the sum of the multiples of 4 from 60 to 100, both inclusive?
A. 240
B. 800
C. 860
D. 880
E. 960
A. 240
B. 800
C. 860
D. 880
E. 960
14 May 2026, 21:19
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How I solved this question-
1- Identifying that multiples of 4 will be in a form of Arithmetic Progression, hence we can use the sum formula.
2- First term is 60 and last term is 100.
3- Number of terms= (100-60)/4 +1= 11 terms (both terms inclusive)
5- Apply the formula= (first term + last terms) * number of terms / 2= (60+100)*11/2= 880
So, 880 that's option D is the answer.
Thanks!
1- Identifying that multiples of 4 will be in a form of Arithmetic Progression, hence we can use the sum formula.
2- First term is 60 and last term is 100.
3- Number of terms= (100-60)/4 +1= 11 terms (both terms inclusive)
5- Apply the formula= (first term + last terms) * number of terms / 2= (60+100)*11/2= 880
So, 880 that's option D is the answer.
Thanks!
14 May 2026, 23:31
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first identify the multiples of 4 from 60 to 100
use the given range to find the first and last multiple of the sequence.
60/4 = 15
100/4 = 25
between this range there are (25-15)+1 = 11 numbers that are multiples of 4.
the multiples are: 60,64, 68,72, ....., 96,100
now rather than adding these num together, we can use arithmetic progression formula,
n/2 [2a + (n-1)d]
here, n= 11 , d=4, a= 60
11/2 [2*60 + (11-1)4]
11/2 [120+40]
11/2[160]
80*11
880
ans D
use the given range to find the first and last multiple of the sequence.
60/4 = 15
100/4 = 25
between this range there are (25-15)+1 = 11 numbers that are multiples of 4.
the multiples are: 60,64, 68,72, ....., 96,100
now rather than adding these num together, we can use arithmetic progression formula,
n/2 [2a + (n-1)d]
here, n= 11 , d=4, a= 60
11/2 [2*60 + (11-1)4]
11/2 [120+40]
11/2[160]
80*11
880
ans D
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