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Question
The 16th term of an AP is 1 more than twice its 8th term. If the 12th term of the AP is 47, then find its nth term ?
Solution
Let a be the first term and d be the common difference of the given A.P.
According to the given question,
16th term of the AP = 2 × 8th term of the AP + 1
i.e., a16 = 2a8 + 1
`a+(16-1)d=2[a+(8-1)d]+5` `thereforea_n=a+(n-1)d`
`rArra+15d=2[a+7b]+5`
`rArra+15d=2a+14d+5`
`rArrd=a+1 .............(1)`
Also, 12th term, a12 = 47
`rArra+(12-1)d=47`
`rArra+11d=47`
`rArra+11(a+1)=47` [Using(1)]
`rArra+11a+11=47`
`rArr12a=36`
`rArra=3`
On Putting the value of a in (1), we get d = 3 + 1 = 4
Thus, nth term of the AP, an= a + (n − 1)d
On putting the respective values of a and d, we get
an= 3 + (n − 1) 4 = 3 + 4n − 4 = 4n − 1
Hence, nth term of the given AP is 4n − 1.
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