Sequences & Series - Arithmetic - the General term.
Test Yourself 1 - Solutions.
| Finding a term | 1. We are given the relationship that
Tn = 33 + 6n. Find T22. Let n = 22. T22 = 33 + 6 × 22 = 165. |
2. Find the 40th term in the sequence 5, 8, 11, ... T1 = 5 and d = 3 T40 = 5 + (40-1) × 3 T40 = 122. |
| 3. An arithmetic sequence has -3 and 5 as two consecutive terms. The 18th term is 101.
Write down the 17th and 20th terms. As the terms are consecutive, the common difference is 8. T18 = 101 so T17 = 101 - 8 = 93. T20 = T18 + 2 × 8 = 117. |
4. The 1st term in an arithmetic sequence is 21 and the common difference is -10.
Find the 17th term. T1 = 21 and d = -10. So T17 = 21 + (17-1) × (-10)
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| Finding a 1st term | 5. An arithmetic sequence has a common difference of 4 and its 27th term is 117.
Find the 1st term. d = 4 and T27 = 117. 117 = T1 + (27-1) × 4 T1 = 13 So the 1st term is 13. |
6. An arithmetic sequence has a common difference of -17 and its 23th term is -352.
Find the 1st term. d = -17 and T23 = -352. -352 = T1 + (23-1) × -17 T1 = 22. So the first term is 22. |
| Finding a difference | 7. The first term in an arithmetic sequence is 4 and the 25th term is 172.
Find the common difference. T1 = 4 and T25 = 172 = 4 + (25-1)d 24d = 168 so d = 7. The common difference is 7. |
8. The first term in an arithmetic sequence is 42 and the 18th term is 33.5.
Find the common difference. T1 = 42 and T18 = 33.5 = 42 + (18-1)d 17d = -8.5 so d = -0.5. The common difference is -0.5. |
| Finding the number of terms | 9. Find the number of terms in the arithmetic sequence 10, 7, 4, ..., -47. T1 = 10 and d = -3 So -47 = 10 + (n-1)×(-3) -60 = -3n n = 20 There are 20 terms in this sequence. |
10. Which term is -372 if it is part of an arithmetic series starting at 240 and having a common difference of -9?
T1 =240 and d = -9 -372 = 240 + (n-1)× (-9) 9n = 621 n = 69. So -372 is the 69th term. |
11. Find the number of multiples of 7 between 300 and 650. To find the first multiple, divide 300 by 7 and round up. So 42.9 becomes 43 to go just past 300. Hence the first term is 43×7 = 301. To find the last multiple, divide 650 by 7 and round down. So 92.9 becomes 92 to go just under 650. Hence the last term is 92×7 = 644. These terms form the first and last terms in the series so: 644 = 301 + (n-1)×7 7n = 350 n = 50. S0 there are 50 multiples of 7 between 300 and 650. Comment.You could subtract the 43 from the 92 AND THEN ADD 1. Are you going to remember that in the exam? Probably not. The slight extra effort of substituting into Tn will avoid you losing marks. |
12. What is the first term in the sequence -8, 5, 18, ... greater than 1250?
T1 = -8 and d = 13. 1250 = -8 + (n-1)× 13. 13n = 1271 n = 97.8 so say 98. T98 = -8 + 97×8 = 1253 To the first term in the sequence greater than 1250 is T98 = 1253.
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| Given 2 terms | 13. If the 6th term of an arithmetic sequence is 17 and the 13th term is 80, what are the first three terms in the sequence?
T6 = T1 + 5d = 17 T13 = T1 + 12d = 80 Subtracting 7d = 63 d = 9 T1 = -28 First three terms are -28, -19, -10. |
14. If the 4th term of an arithmetic sequence is 27 and the 7th term is 12, what is the common difference for the sequence?
T4 = T1 + 3d = 27 T7 = T1 + 6d = 12 Subtracting 3d = -15 d = -5 {T1 = 42} |
| 15. For a particular arithmetic sequence, three times the 3rd term equals the 9th term.
What could be the first three terms of a possible sequence with these properties? 3 × T3 = T1 + 8d 3(T1 + 2d) = T1 + 8d 2T1 = 2d T1 = d Let d = 2 then T1 = 2 So a possible sequence would be |
16. If the 2nd term of an arithmetic sequence is 7 and the 7th term is 52, find the value of the first term greater than 1,000.
T2 = T1 + d = 7 T7 = T1 + 6d = 52 5d = 45 d = 9 T1 = -2 Let 1000 = -2 + (n-1)9 1011 = 9n n = 112.3 - so say 113 to be greater. T113 = -2 + (113-1) × 9 = 1006 So the first term greater than 1000 is T113 = 1006. |
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| Miscellaneous | 17. The terms x-2, 2x-1, 4x-4 form an arithmetic sequence.
Find the first three terms of the sequence. As it is an arithmetic sequence, write expressions for the common difference and equate them. So (2x-1) - (x-2) = (4x-4) - (2x-1) x + 1 = 2x-3 x = 4. Hence the terms are 2, 7, 12. (a 1st term of 2 and a common difference of 5). |
18. Find the number of multiples of 13 between 350 and 730.
To find the first multiple greater than 350, divide 350 by 13 (=26.9) and round the answer up to 27. The last multiple is found by dividing 730 by 13 (=56.2) and rounding down to obtain 56 so 56 × 13 = 728. So, in the sequence, the 1st term is 351 and the last term is 728. 728 = 351 + (n-1)13 13n = 390 n = 30 There are 30 multiples of 13 between 350 and 730. |
| 19. The sum of the 2nd and 5th terms of an arithmetic series is 32 while the sum of the 3rd and 8th terms is 48.
(i) Find the 1st term and the common difference. (ii) find the 10th term. (i) T2 + T5 = T1 + d + T1 + 4d = 32 2T1 + 5d = 32 T3 + T8 = T1 + 2d + T1 + 7d = 48. 2T1 + 9d = 48. Subtracting: 4d = 16 d = 4 T1 = 6 So the 1st term is 6 and the common difference is 4. (ii) T10 = 6 + 9 × 4 = 42 So the 10th term is 42. |
20. In an arithmetic sequence, the 10th term is double the 6th term. The square of the 3rd term is equal to the 6th term.
Find the first three terms in the sequence. T10 = 2 × T6 T1 + 9d = 2(T1 + 5d) T1 = -d (T3)2 = T6 (T1 + 2d)2 = T1 + 5d Substituting for T1 d2 = 4d d = 0 (not possible) or d = 4 ∴T1 = -4 First three terms are -4, 0, 4. |