The 15^th term of an AP is 3 more than twice its 7^th term. If the 10^th term of the AP is 41, then find its n^th term.

Let 15^th term of the AP series be a₁₅, 7^th term be a₇, 10^th term be a₁₀ and n^thterm be a_n.

According to the question,

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

a₁₅ = 3 + 2a₇ …(i)

and the a₁₀ = 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.

∴, a₁₅ = a + (15 – 1)d = a + 14d. Similarly, a₇ = a + 6d and a₁₀ = 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.