The 17th term of an A.P. is 5 more than twice its 8th term.
Question:

The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43. find the nth term.

Solution:

Let a be the first term and d be the common difference.

We know that, nth term = an a + (n − 1)d

According to the question,

a17 = 5 + 2a8
⇒ a + (17 − 1)d =  5 + 2(a + (8 − 1)d)
⇒ a + 16d =  5 + 2a + 14d
⇒ 16d − 14=  5 + 2a − a
⇒ 2=  5 + a
⇒ a = 2− 5    …. (1)

Also, a11 = 43
⇒ a + (11 − 1)d = 43
⇒ a + 10d = 43   ….(2)

On substituting the values of (1) in (2), we get
2− 5 + 10d = 43
⇒ 12= 5 + 43
⇒ 12= 48
⇒ = 4
⇒ a = 2 × 4 − 5     [From (1)]
⇒ a = 3

∴ an a + (n − 1)d
        =
 3 + (n − 1)4
        = 3 + 4n − 4
        = 4n − 1

Thus, the nth term of the given A.P. is  4n − 1.

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