Find out which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.10+10+25,10+26,10+27,.....
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Solution
Given sequence is,
10+10+25,10+26,10+27...…
first term of this A.P is a1=10+10+25=52
second term of this A.P is a2=10+26=74
third term of this A.P is a3=10+27=138
the condition for an sequence to be an A.P is their must be a common difference (i.e.,d=an+1−an)
putting n=1 in above equation
d=a2−a1=74−52=22
putting n=2 in above equation
d=a3−a2=138−74=64
as we can see we get two values of d , but in an A.P their must be a single value of d which means for this sequence we can't define a common difference