8. whose fifth term is 14 and 12th term is 0.
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Step-by-step explanation:
5th term=14
=> a+4d=14 --------(1)
12th term=0
=>a+11d=0 --------(2)
subtracting (1) and (2)
a+11d=0
a+4d=14
7d= -14
d= -7
substituting value of d in (1)
a+4(-7)=14
a=14+28
=>a=42
therefore the AP is 42, 42+(-7), 42+2(-7)...
=> AP= 42, 35, 28...
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