18619539
| Question | Answer |
| The common difference of an arithmetic sequence is defined by.... | d |
| The first of an arithmetic sequence is given by.... | a |
| How do we find any term of an arithmetic sequence? | a+(n-1)d |
| In general, an arithmetic sequence looks like: | a , a+d , a+2d , a+3d ...... , a+(n-1)d |
| If a=7 and d=5 , find the 10th term of the arithmetic sequence | 7+(10-1)5 = 52 |
| The 5th term of an arithmetic sequence is 50 and the 8th is 92. Find a and d. | a + (5-1)d = 50 , a + (8-1)d = 92 a+4d=50 a+7d=92 Solve simultaneously for a=-6 and d=14 |
| What is an arithmetic series? | The sum of the terms in an arithmetic sequence |
| In general, what does an arithmetic series look like? | Sn = a + (a+d) + (a+2d) + .... + a+(n-1)d |
| How do we find the sum of an arithmetic series? | Sn = 1/2 n(2a+(n-1)d) OR Sn = 1/2 n(a+l) (a= first term, l=last term) |
| The arithmetic series: 51+58+65+72+...+1444 has 200 terms. Find the 101st term and also the sum of the last 100 terms | a=51 , d=7 u101 = 51 + (101-1)7 = 751 Last 100 terms = S200 - S100 , so: 0.5(200)(2x51+(200-1)7) - 0.5(100)(2x51+(100-1)7 = 109750 |
| What is another way we could calculate the sum of the last 100 terms for the question above? | Use u101 = 751 ; a=751 , d=7 , n=100 So: 0.5(100)(2x751 + (100-1)7) = 109750 |
| The 25th term of an arithmetic series is 38. The sum of the first 40 terms is 1250. Show the common difference is 1.5 and the first term is 2. | a+(25-1)d = 38 , a+24d = 38 0.5(40)(2a+(40-1)d) = 1250 , so 4a+78d=125 Solve these 2 equations simultaneously for a=2 and d=3/2 |
| For the values above, find the number of terms in the series which are less than 100 | 2 + (n-1)1.5 < 100 , rearrange for n<199/3 199/3 = 66 + 1/3 , so the number of terms less than 100 = 66 (ignore 1/3 as not integer) |
| The nth term of an arithmetic series is 8n-6 , find the sum of the first 20 terms. | First term = 8(1)-6 = 2 Last term = 8(20)-6 = 154 Therefore: 0.5(20)(2+154) = 1560 |
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