How do you write the first five terms of the arithmetic sequence given #a_1=72, a_(k+1)=a_k-6# and find the common difference and write the nth term of the sequence as a function of n?

Answer 1

First five terms are # 72 , 66 ,60 , 54 and 48 #. The #n# th term of the sequence is #T_n= 72-6(n-1)#

#a_1=72, a_(k+1)=a_k-6 :. a_2 =a_1-6=72-6=66#
# a_3 =a_2-6=66-6=60 ; a_4 =a_3-6=60-6=54#
#a_5 =a_4-6=54-6=48 :. # First five terms are
# 72 , 66 ,60 , 54 and 48 #. This is arithmatic progression
series of which common difference is #d = 66-72=-6 #
similarly #d = 60-66 = -6 # . The #n# th term of the sequence
is #T_n = a_1 +(n-1) d or T_n= 72 + (n-1)* -6# or
#T_n= 72-6(n-1)# , Check # T_5 = 72 -6(5-1) =48# [Ans]
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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