Answer :

Let 15^{th} term of the AP series be a_{15}, 7^{th} term be a_{7}, 10^{th} term be a_{10} and n^{th}term be a_{n}.

According to the question,

15^{th} term of an AP is 3 more than twice its 7^{th} term.

a_{15} = 3 + 2a_{7} …(i)

and the a_{10} = 41 [∵, 10^{th} term is 41] …(ii)

Now a_{n} is given by a_{n} = a + (n – 1)d, where a is first term of the AP, n is total number in the series and d is common difference between adjacent numbers in the series.

∴, a_{15} = a + (15 – 1)d = a + 14d. Similarly, a_{7} = a + 6d and a_{10} = a + 9d

From equation (i),

(a + 14d) = 3 + 2(a + 6d)

⇒ a + 14d = 3 + 2a + 12d

⇒ 14d – 12d = 2a – a + 3

⇒ 2d = a + 3

⇒ 2d – a = 3 …(iii)

From equation (ii),

(a + 9d) = 41

9d + a = 41 …(iv)

Adding equations (iii) and (iv), we get

(2d – a) + (9d + a) = 3 + 41

⇒ 11d = 44

⇒ d = 44/11 = 4

⇒ d = 4

Putting d = 4 in equation (iii), we get

2d – a = 3

⇒ a = 2d – 3

⇒ a = 2(4) – 3 = 8 – 3 = 5

Substituting a = 5 and d= 4 in a_{n} = a + (n – 1)d,

a_{n} = 5 + (n – 1)(4) = 5 + 4n – 4 = 4n + 1

Thus, we have got n^{th} term, that is, 4n + 1.

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