In an arithemetic sequence 3rd term 37 and 7th term is 73 find 30th term,
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Answer:
A3=37,A7=73
a+2d=37,a+6d=73
on solving both equations
4d=36
d=9
substitute d=9 in a+2d=37
a+18=37
a=19
A30=a+29d
A30=19+29(9)
A30=19+261
A30=280
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