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16 Sep 2014, 01:21
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What is the value of \(1 + 2 + 3 + ... + 47 + 48 + 49\) ?
A. 1168
B. 1176
C. 1195
D. 1215
E. 1225
A. 1168
B. 1176
C. 1195
D. 1215
E. 1225
16 Sep 2014, 01:21
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Bookmarks
Official Solution:
What is the value of \(1 + 2 + 3 + ... + 47 + 48 + 49\) ?
A. 1168
B. 1176
C. 1195
D. 1215
E. 1225
To find the sum of an evenly spaced set, we multiply the average (arithmetic mean) by the number of terms.
The average of an evenly spaced set is the average of the first and the last terms, hence it's \(\frac{1 +49}{2}=25\). Therefore, the sum is \(25*49 = 1225\).
Answer: E
What is the value of \(1 + 2 + 3 + ... + 47 + 48 + 49\) ?
A. 1168
B. 1176
C. 1195
D. 1215
E. 1225
To find the sum of an evenly spaced set, we multiply the average (arithmetic mean) by the number of terms.
The average of an evenly spaced set is the average of the first and the last terms, hence it's \(\frac{1 +49}{2}=25\). Therefore, the sum is \(25*49 = 1225\).
Answer: E
06 Feb 2018, 10:31
Kudos
Bookmarks
Hi,
1,2,3..49 is an AP series.
So for that we can take the sum as averageXno of integers
Here average = (first term+last term)/2--> 1+49/2=25
Sum =25x49 = 1225
1,2,3..49 is an AP series.
So for that we can take the sum as averageXno of integers
Here average = (first term+last term)/2--> 1+49/2=25
Sum =25x49 = 1225
