Question

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 ?
A) 14th element
B) 9th element
C) 12th element
D) 7th element

Answer 1

Answer:  The correct option is (C) 12th element.

Step-by-step explanation:  Given that the sum of the 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression.

We are to find the element of the series that should be necessarily zero.

Let a and d be the first term and common difference of the given arithmetic progression.

Then, the n-th term o the A.P. will be

$a_n=a+(n-1)d.$

According to the given information, we have

$a_3+a_{15}=a_6+a_{11}+a_{13}\\\\\Rightarrow a+(3-1)d+a+(15-1)d+a+(6-1)d+a+(11-1)d+a+(13-1)d\\\\\Rightarrow 2a+2d+14d=3a+5d+10d+12d\\\\\Rightarrow 2a+16d=3a+27d\\\\\Rightarrow 3a-2a+27d-16d=0\\\\\Rightarrow a+11d=0\\\\\Rightarrow a+(12-1)d=0\\\\\Rightarrow a_{12}=0.$

Thus, the 12th element of the A.P must be zero.

Option (C) is CORRECT.

Answer 2

ANSWER

Let the first term of AP be a and difference be d

Then third term will be =a+2d

${15}^{th} \: will \: be = a + 14d$

${6}^{th} \: will \: be = a + 5d$

$1 {1}^{th} \: will \: be = a + 10d$

$1 {3}^{th} will \: be = a + 12d$

$then \: the \: eq. \: will \: be$

$a + 2d + a + 14d = a + 5d + a + 10d + a + 12d$

$= > 2a + 16d = 3a + 27d$

$= > a + 11d = 0$

$we \: understand \: a + 11d \: will \: be \: the \: 1 {2}^{th} \: term \: of \: arithmetic \: progression.$

$so, \: CORRECT \: answer \: is \: {\boxed {\pink{12}}}$

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