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Question
The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
Solution
In the given problem, we are given 6th and 17th term of an A.P.
We need to find the 40th term
Here
`a_6 = 19`
`a_17 = 41`
Now we will find `a_6` and `a_17` using the formula `a_n = a + (n - 1)d`
So
`a_6 = a + (6 - 1)d`
19 = a + 5d .......(1)
Also
`a_17 = a + (17 - 1)d`
41 = a + 16d .....(2)
So to solve for a and d
On subtracting (1) from (2) we get
`a + 16d - a - 5d = 41 - 19`
11d = 22
`d = 22/11`
d = 2 .....(3)
Substituting (3) in (1), we get
`19 = a + 5(2)`
19 - 10 = a
a = 9
Thus
a = 9
d = 2
n = 40
Substituting the above values in the formula `a_n = a + (n -1)d`
`a_40 = 9 + (40 - 1`)2`
`a_40 = 9 + 80 - 2`
`a_40 = 87`
Therefore `a_40 = 87`
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