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by harsh.champ » Thu Feb 04, 2010 1:40 pm
The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?


(1)1st
(2)9th
(3)12th
(4)6th
(5)None of the above
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by harsh.champ » Thu Feb 04, 2010 3:32 pm
I solved this question using the following method:-
Let the first term be a and the common difference be n. [a,a+n,a+2n...]
(a+2n)+(a+14n) = (a+5n)+(a+10n)+(a+12n)
Hence, 2a +16n= 3a+27n
=>a=-11n
=>n=-a/11
putting the value of n in the question,the series becomes a,a-a/11,a-2a/11......a-11a/11
Hence it will be the 12th term.

Is there any shortcut method to solve such a type of question??
Last edited by harsh.champ on Thu Feb 04, 2010 4:21 pm, edited 2 times in total.
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by deepak_free » Thu Feb 04, 2010 3:41 pm
I think its 3

(a+2n)+(a+14n) = (a+5n)+(a+10n)+(a+12n)

a+11n=0

and this is nothing but 12th element
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by ajith » Thu Feb 04, 2010 4:14 pm
harsh.champ wrote:The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?


(1)1st
(2)9th
(3)12th
(4)6th
(5)None of the above
nth term for an arithmetic progression = a+(n-1)*d

3rd term = a+2d
15th term = a+14d

sum = 2a+16d

6th term =a+5d
11th term = a+10d
13th term = a+12d

sum = 3a+27d

now 3a+27d = 2a+16d
a+11d =0

12th term = a+11d
12th term =0
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