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A common arithmetic progression is

a, a+d, a+2d, a+3d, a+4d,.....

where 'a' is the first element and 'd' is the common difference

As per the question,

the sum of 3rd and 15th elements is equal to the sum of 6th, eleventh and 13th elements, deriving from the formula

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

2a+16d=3a+27d

2a+16d-3a-27d=0

a+11d=0

So the 12th element of the series is ZERO

a, a+d, a+2d, a+3d, a+4d,.....

where 'a' is the first element and 'd' is the common difference

As per the question,

the sum of 3rd and 15th elements is equal to the sum of 6th, eleventh and 13th elements, deriving from the formula

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

2a+16d=3a+27d

2a+16d-3a-27d=0

a+11d=0

So the 12th element of the series is ZERO