Aryan got − 1 0 marks in his first exam and 1 5 marks in his 1 5 th exam.
If all his marks follow an arithmetic progression with a positive common difference, in which exam did he get zero marks?
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I think you made a typo error, the common difference must be 1 4 5 .
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15 = -10 + (15-1)d.. How did you get d to be 5/14 here?
Yes the common difference is 5/14
Hence
0 = a + (n - 1 ) d 0 = 10 + ( n - 1 ) 5/14 0 = 10 + 5n/14 - 5/14
We take fraction LCM
0 = 140 + 5n - 5
Like terms
0 = 135 + 5n
135 = -5n
Divide both side by 5
n = -27
The common difference is 25/14 so if 0 was a term in the AP then the solution to 0 = -10 + (n - 1)25/14 would be a natural number. As it is 6.6 we know that 0 is not a term in the AP.
Since the first exam is − 1 0 marks and the 1 5 t h exam is 1 5 marks and the common difference is positive, z e r o m a r k s is not possible.
May I bring to your kind notice that the marks in the first exam is -10, not 10.
华文:第一个是十分,第十五个是十五分。变多了。第一个是十分的话怎么可能有零分? ... English:Aryan got 10 marks in his first exam and 15 marks in his 15th exam. 15th exam is more than 1st exam. If the 1st exam is already 10, how is it possible that in one of the exams he only got 10 marks? ... Melayu: Yang pertama adalah sepuluh mata dan yang kelima belas adalah lima belas mata. Lebih banyak lagi. Bagaimana boleh ada titik sifar jika yang pertama adalah sepuluh?
It does not require calculation. The problem statement says common difference is positive. If the sequence has started from a positive number, and C.D is positive, then no member of the sequence could be zero.
It doesn't start from a positive number. The initial value is -10
Yes the common difference is 5/14
Hence
0 = a + (n - 1 ) d 0 = 10 + ( n - 1 ) 5/14 0 = 10 + 5n/14 - 5/14
We take fraction LCM
0 = 140 + 5n - 5
Like terms
0 = 135 + 5n
135 = -5n
Divide both side by 5
n = -27
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The common ratio is d= 25/14 according to which 0 shall never come.