Question

The 16^th term of an AP is 1 more than twice its 8^th term. If the 12^th term of the AP is 47, then find its n^th term ?

Answer 1

Let a be the first term and d be the common difference of the given A.P.

According to the given question,

16^th term of the AP = 2 × 8^th 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, 12^th 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, n^th term of the AP, a_n= a + (n − 1)d

On putting the respective values of a and d, we get

a_n= 3 + (n − 1) 4 = 3 + 4n − 4 = 4n − 1

Hence, n^th term of the given AP is 4n − 1.