find the m term of an arithmetic progression whose 12 term exceeds the 5 term by 14 and the sum of both term is 36
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Answer:
mth term=1+2m
Step-by-step explanation:
12th term=a+11d
5th term=a+4d
acc to ques.
(a+11d)-(a+4d)=14
7d=14
d=2
also (a+11d)+(a+4d)=36
putting the value of d and solving for a we get
a=3
mth term=a+(m-1)d
=3+(m-1)2
mth term=1+2m
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