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Solutions to exercises on finding terms and common differences in arithmetic sequences. It includes various examples and justifications for the answers.
Typology: Summaries
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Supported by UBC Teaching and Learning Enhancement Fund 2012-
D e p a r t m e n t o f C u r r i c u l u m a n d P e d a g o g y
a place of mind^ F A C U L T Y^ O F^ E D U C A T I O N
Answer: B
Justification: The sequence is called an arithmetic sequence because the difference between any two consecutive terms is 2 (for example 6 – 4 = 2). This is known as the common difference. The next term in the sequence can be found by adding the common difference to the last term:
5 th^ term
+2 +2 +
8 th^ term
A. a 8 = a 5 + 3d B. a 8 = a 5 + 3a 1 C. a 8 = a 5 + 8d D. a 8 = a 5 + 8a 1 E. Cannot be determined
Consider the following sequence of numbers:
a 1 , a 2 , a 3 , a 4 , a 5 , ...
where an is the nth^ term of the sequence. The common difference between two consecutive terms is d. What is a 8 , in terms of a 5 and d?
A. a 8 = 8a 1 B. a 8 = a 1 + 6d C. a 8 = a 1 + 7d D. a 8 = a 1 + 8d E. Cannot be determined
Consider the following sequence of numbers:
a 1 , a 2 , a 3 , a 4 , a 5 , ...
where an is the nth^ term of the sequence. The common difference between two consecutive terms is d. What is a 8 , in terms of a 1 and d?
Answer: C
Justification: The next term in the sequence can be found by adding the common difference to the previous term. Starting at the first term, the common difference must be added 7 times to reach the 8th^ term:
a 1 , a 2 , a 3 , a 4 , a 5 , a 6 , a 7 , a 8
Note how we do not add 8 times the common difference to reach the 8th^ term if we are starting at the first term.
Answer: D
Justification: Consider the value of the first few terms: a 1 = a 1 + 0d a 2 = a 1 + 1d a 3 = a 1 + 2d a 4 = a 1 + 3d ⋮ an = a 1 + (n-1)d
Notice that the common difference is added to a 1 (n-1) times, not n times. This is because the common difference is not added to a 1 to get the first term. Also note that the first term remains fixed and we do not add multiples of it to find later terms.
A. a 21 = 6 + 20(5) B. a 21 = 21 + 20(5) C. a 21 = 21 + 21(5) D. a 21 = 21 – 20(5) E. a 21 = 21 – 21(5)
__, __, __, 6, 1, ...
Press for hint
Hint: Find the value of the common difference and the first term. an = a 1 + (n-1)d
A. 998 B. 999 C. 1000 D. 1001 E. 1002
How many numbers are there between 23 and 1023 inclusive (including the numbers 23 and 1023)?
Press for hint
Hint: Consider an arithmetic sequence with a 1 = 23, an = 1023, and d = 1 an = a 1 + (n-1)d
Answer: D
Justification: The answer is not just 1023 – 23 = 1000. Imagine if we wanted to find the number of terms between 1 and 10. The formula above will give 10 – 1 = 9, which is incorrect.
Consider an arithmetic sequence with a 1 = 23, and an = 1023. The common difference (d) for consecutive numbers is 1. Solving for n, we can find the term number of 1023:
Since 1023 is the 1001th^ term in the sequence starting at 23, there are 1001 numbers between 23 and 1023.
n 1001
n- 1 1023 23
1023 23 (n 1)
a (^) n a 1 (n 1)d
Answer: D
Justification:
(Method 2): Using the formulas, a 19 and a 30 in terms of a 1 is given by: a 19 = a 1 + 18d a 30 = a 1 + 29d Subtracting a 30 from a 19 gives:
11
30 d
30 11d
a 30 a 19 11d
(Method 1): To get to a 30 from a 19 , 11 times the common difference must be added to a 19 :
11
30 d
30 11d
a a 11d
a a 11d
30 19
30 19
A. a 1 = 10; d = 2 B. a 1 = 15; d = - C. a 11 = 30; a 12 = 20 D. a 20 = 40; d = 2 E. a 20 = 40; d = -
The statements A through E shown below each describe an arithmetic sequence. In which of the arithmetic sequences is the value of a 10 the largest?
