Question

The 17 term of an AP is 5 more than twice its 8th term. If the 11" term of the AP is 43, find the n th term.​

Answer 1

Given :-

◉ 17th term of an AP is 5 more than twice its 8th term.

We know, nth term of an AP is in the form:

aₙ = a + (n - 1)d

So,

17th term = a + 16d

8th term = a + 7d

◉ 11th term of the AP is 43.

∴ a + 10d = 43 ...(1)

To Find :-

◉ nth term or aₙ

Solution :-

Given in the question,

⇒ 17th term = 2(8th term) + 5

⇒ a + 16d = 2a + 14d + 5

⇒ 2a - a + 14d - 16d + 5 = 0

⇒ a = 2d - 5 ...(2)

Now that we have got the first term in terms of d so we can now find the common difference of the AP by substituting a = 2d - 5 in (1)

⇒ 2d - 5 + 10d = 43

⇒ 12d = 48

⇒ d = 4

Again, Substitute d = 4 in (2),

⇒ a = 2×4 - 5

⇒ a = 8 - 5

⇒ a = 3

So, we have the common difference and the first term of the AP, let's find the aₙ ( nth-term of the AP ) which would be our answer.

⇒ aₙ = a + (n - 1)d

⇒ aₙ = 3 + (n - 1)4

⇒ aₙ = 3 + 4n - 4

⇒ aₙ = 4n - 1

Hence, Nth-term of the AP is 4n - 1

Answer 2

The 17 term of an AP is 5 more than twice its 8th term.

→ a + (17 - 1)d = 2(a + (8-1)d) + 5

→ a + 16d = 2a + 14d + 5

→ 2a - a + 14d - 16d + 5 = 0

→ a = 2d - 5

11th term of the AP is 43.

→ a + (11 - 1)d = 43

→ a + 10d = 43

→ a = 43 - 10d

On comparing both we get,

→ 2d - 5 = 43 - 10d

→ 2d + 10d = 43 + 5

→ 12d = 48

→ d = 4

Substitute value of d in one of the equation,

→ a = 2(4) - 5

→ a = 8 - 5

→  a = 3

We have to find the nth term of the AP. So,

→ an = a + (n - 1)d

→ an = 3 + (n - 1)4

→ an = 3 + 4n - 4

→ an = 4n - 1

Hence the nth term of an AP is 4n - 1.