Question

8. whose fifth term is 14 and 12th term is 0.​

Answer 1

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...