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Question
Find the arithmetic progression whose third term is 16 and the seventh term exceeds its fifth term by 12.
Solution
Here, let us take the first term of the A.P as a and the common difference of the A.P as d
Now, as we know,
`a_n = a + (n - 1)d`
So for the 3rd term (n = 3)
`a_3 = a + (3 - 1)d`
16 = a + 2d
a = 16 - 2d ......(1)
Also for 5th term (n = 5)
`a_5 = a + (5 - 1)d`
= a + 4d
For 7th term (n = 7)
`a_7 = a + (7 - 1)d`
=a + 6d
Now we are given
`a_7 = 12 + a_5`
a + 6d = 12 + a + 4d
6d - 4d = 12
2d = 12
d = 6
Substituting the value of d in (1), we get,
`a = 16 - 2(6)`
= 16 - 12
= 4
So the first term is 4 and the common difference is 6
Therefore the A.P is 4, 10 , 16, 22,....
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