101,99,97,95,..., Determine if the given sequences represent an AP, assuming that the pattern continues. If it is an AP, find the nth term.
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101, 99, 97, 95........
In this sequence, each term is obtained by adding -2 to the immediately preceding term.
Hence the sequence is an A.P
a = 101, d = -2
The n th term the sequence is
tn = a +(n-1)d
tn = 101+(n-1)(-2)
tn = 101 -2n+2
tn = 103-2n
I hope this answer helps you
Answered by
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we know, 
In the above sequence,
a = 101;
d₁ = a₂–a₁ = 99–101 = –2
d₂ = a₃–a₂ = 97–99 = –2
d₃ = a₄–a₃ = 95–97 = –2
⇒ As in A.P the difference between the 2 terms is always constant
The difference in sequence is same and comes to be (–2).
∴ The above sequence is A.P
The nth term of A.P is aₙ = a + (n–1)d
aₙ = a + (n–1)d = 101 + (n-1)(–2)
= 101–2n + 2
= 103–2n
In the above sequence,
a = 101;
d₁ = a₂–a₁ = 99–101 = –2
d₂ = a₃–a₂ = 97–99 = –2
d₃ = a₄–a₃ = 95–97 = –2
⇒ As in A.P the difference between the 2 terms is always constant
The difference in sequence is same and comes to be (–2).
∴ The above sequence is A.P
The nth term of A.P is aₙ = a + (n–1)d
aₙ = a + (n–1)d = 101 + (n-1)(–2)
= 101–2n + 2
= 103–2n
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