Math, asked by aubreeosborne, 1 year ago

An arithmetic sequence is defined by the recursive formula t1 = 44, tn + 1 = tn + 16, where n ∈N and n ≥ 1. Which is the general term of the sequence? A) tn = 44 + (n - 1)16, where n ∈N and n ≥ 1 B) tn = 44 + (n + 1)16, where n ∈N and n ≥ 1 C) tn = 44 + (n - 2)16, where n ∈N and n ≥ 1 D) tn + 1 = 44 + (n - 1)16, where n ∈N and n ≥ 0

Answers

Answered by nathswami619
1

Answer:

tn=a+n-1 Xd

Step-by-step explanation:

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Answered by Pikaachu
4

Answer:

(A)

Step-by-step explanation:

t_{n + 1} = t_{n} + 16

 =  > t_{n + 1} - t_{n} = 16

The above equation clearly reminds us of an A.P.

Elegant Soln :

Define an A.P.

 t_{n} = a + (n - 1)d

-> 'a' being the first term of the sequence can be switched as :

t_{1} = a + (1 - 1)d = a

Further,

t_{n + 1} - t_{n} = (a + (n + 1 - 1)d) -  (a + (n - 1)d)

t_{n + 1}  - t_{n} = d = 16

 =  > t_{n} = 44 + (n - 1)16, n \geqslant 1, \: n  \epsilon N

Basic Solving :

t_{n + 1} - t_{n} = 16

 =  >  t_{n } - t_{n - 1} = 16

.

.

.

  =  > t_{2} - t_{1} = 16

Adding all equations above and cancelling adjacent terms, we get :

 t_{n + 1} - t_{1} = 16n

 =  > t_{n}  = t_{1} + (n - 1)16

So, hopefully you get the answer.

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