Find the mth term of an Arithmetic progression whose 12th
term exceeds the 5th term by 14 and the sum of both terms
is 36
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2
Answer:
a12=a5+14
a+11d=a+4d+14
11d-4d=14
7d=14
d=2 now
a12+a5=36
a+11d+a+4d=36
2a=36-15d. where d=2
2a=36-15×2
2a=6
a=3
now mth term equal to ,,,amth=a+(m-1)d
amth=3 +(m-1)2
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Answer:
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