Question 1149133: if 4 arithmetic means are inserted between 1 & 36 the sum of the resulting terms is 148 .find the common difference and n . thank you
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52855)   (Show Source):
You can put this solution on YOUR website! .
If 4 arithmetic means are inserted between 1 and 36, it means that you have an arithmetic progression with 6 terms:
the first term is 1 and the last, 6-th term is 36.
There are 5 gaps in the number line between these 6 terms of the AP.
The gaps are of the same length, which is, hence,  =  = 7.
Thus, the common difference is exactly the value of the gap, i.e. 7 units.
Thus the AP is 1, 8, 15, 22, 29, 36.
ANSWER. n = 6, which is MORE THAN OBVIOUS.
Common difference is 7.
Solved.
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Notice. The info about the sum (148) is not necessary and excessive.
It is NOT USED in the solution.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
The given information is inconsistent -- i.e., self-contradictory.
When 4 arithmetic means are inserted between 1 and 36, the difference of 35 is divided into 5 equal intervals.
35/5 = 7; the common difference is 7.
The 4 arithmetic means are 8, 15, 22, and 29.
The sum of the 4 arithmetic means is 74.
The sum of all 6 terms is 111.
There is no sum of 148 anywhere....
Note if the number of arithmetic means is 6, then the problem is valid.
6 arithmetic means --> 35 is divided into 7 intervals.
35/7 = 5; the common difference is 5.
The 6 arithmetic means are 6, 11, 16, 21, 26, 31; their sum is 111.
The sum of the terms including the 1 and 36 is 148.
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