In the 3rd and the 9th term of an ap are 4 and -8 respectively , which term of this ap is zero
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It is given that 3rd and ninth term of AP are 4 and -8 respectively.
It means ᵃ₃=4 and a₉=−8 where, a₃ and a9 are third and ninth terms respectively.
Using formula an=a+(n−1)d, to find nth term of arithmetic progression, we get
4 = a + (3-1)d And, -8 = a + (9-1)d
⇒4=a+2d And, −8=a+8d
These are equations in two variables. Lets solve them using method of substitution.
Using equation
4=a+2d,
we can say that
a=4−2d.
Putting value of a in other equation
−8=a+8d,
we get
−8=4−2d+8d
⇒−12=6d
⇒d=−126=−2
Putting value of d in equation
−8=a+8d,
we get
−8=a+8(−2)
⇒−8=a−16
⇒a=8
Therefore, first term =a=8 and
Common Difference =d=−2
We want to know which term is equal to zero.
Using formula an=a+(n−1)d, to find nth term of arithmetic progression, we get
0=8+(n−1)(−2)
⇒0=8−2n+2
⇒0=10−2n
⇒2n=10
⇒n=102=5
Therefore, 5th term is equal to 0.
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