Advertisements
Advertisements
Question
Find the next five terms of the following sequences given by:
`a_1 = -1, a_n = (a_n - 1)/n, n>= 2`
Solution
`a_1 = -1, a_n = (a_n - 1)/n, n>= 2`
Here, we are given that `n >= 2`
So, the next five terms of this A.P would be `a_2, a_3, a_4` and `a_6`
Now `a_1 = 1` ......(1)
So to find the `a_2` term we use n = 2 we get
`a_2 = (a_2 -1)/2`
`a_2 = a_1/2`
`a_2 = (-1)/2` (Using 1)
`a_2 = (-1)/2` .....(2)
For `a_3` using n = 3 we get
`a_3 = (a_3 -1)/3`
` a_3 = a_2/3`
`a_3 = (-1/2)/3` ....(3)
For `a_4` using n = 4 we get
`a_4 = (a_4 -1)/4`
``a_4 = a_3/4``
`a_4 = (-1/6)/4` (Using 3)
`a_4 = (-1)/24`...(4)
For `a_5` using n = 5 we get
`a_5 = (a_5 - 1)/5`
`a_5 = (-1/24)/5` (using 4)
`a_5 = (-1)/120` .....(5)
For `a_6` using n = 6 we get
`a_6 = (a_6 - 1)/6`
`a_6 = a_5/6`
`a_6 = (-1/120)/6` (Using 5)
`a_6 = (-1)/720`
Thereore the next five term of the given A.P are `a_2 = (-1)/3, a_3 = (-1)/6,a_4 =(-1)/24,a_5 = (-1)/120, a_6 = (-1)/720`
APPEARS IN
RELATED QUESTIONS
Write the first five terms of the following sequences whose nth terms are:
an = n2 − n + 1
A man saved Rs 16500 in ten years. In each year after the first, he saved Rs 100 more than he did in the preceding year. How much did he save in the first year?
Find:
(ii) the 35th term of AP 20,17,14,11,..........
How many terms are there in the AP 6,1 0, 14, 18, ….., 174?
If sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is –3 then find the 10th term.
In an AP, if d = –4, n = 7, an = 4, then a is ______.
The (n - 1)th term of an A.P. is given by 7, 12, 17, 22,… is ______.
Is 0 a term of the AP: 31, 28, 25, ...? Justify your answer.
Justify whether it is true to say that the following are the nth terms of an AP.
3n2 + 5
Fill in the blank in the following table, given that a is the first term, d the common difference, and an nth term of the AP:
a | d | n | an |
______ | -3 | 18 | -5 |