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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 n th term.

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Given:

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

To find:

the nth term.

Solution:

Given, a17 = 5 + 2(a8) ......(i)

And a11 = 43

As an = a + (n -1) d

For n = 11,

a + 10d = 43a = 43 - 10d ......(ii)

For n = 8, a8 = a + 7d

For n =17,

a17 = a + 16d

Put the value of a from (ii)

⇒ a17= 43 – 10d + 16d

⇒ a17 = 43 + 6d

Putting the value of a8 and a17 in (i), we get

⇒ 43 + 6d = 5 + 2(a + 7d)

⇒ 43 + 6d = 5 + 2a + 14d

⇒ 43 – 5 = 2a + 14d – 6d

⇒ 38 = 2a + 8d

⇒ 38 = 2(43 – 10d) + 8d [from (ii)]

⇒ 38 = 86 – 20d + 8d

⇒ 38 = 86 – 12d

⇒ 12d = 86 – 38

⇒ 12d = 48

⇒ d = 4

From (ii),

a = 43 – 10d

⇒ 43 – (10 × 4) = 43 – 40 = 3

We know, nth term of A.P., an = a + (n -1) d

an = 3 + (n – 1) 4

an = 3 + 4n – 4 an = 4n – 1

Hence, nth term is 4n – 1.

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