Question

The sum of 4th & 16th term of an arithmetic progression is equal to sum of 7th, 12th and 14th elements of same progression. Then which element of the series should necessarily be equal to zero.
A) 1st
B) 13thterm
C) 9th
D) 10thterm

Answer 1

Answer:  B) 13th term

Step-by-step explanation:

 given

t4 + t16 = t7 + t12 + t14

a + 3d + a + 15d = a + 6d + a + 11d + a + 13d                  [ tn = a + ( n - 1 ) d ]

2a + 18d = 3a + 30d

3a + 30d - 2a - 18d = 0

a + 12d = 0

t13 = 0

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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