the sum of the 1st and 17th terms of an arithmetic sequence is 40.the sum of its 1st and 18th term is 43( question no 1.) what is its common difference? (question no.2) what is the sum of its 7th and 11th terms?(question no.3)find it's 9th term?​

Sum of 1st and 17th terms of an arithmetic sequence is 40.

Sum of it's 1st and 18th terms of an arithmetic sequence is 43.

As we know that,

General terms of an ap.

⇒ Tₙ = a + (n - 1)d.

Using this formula in this question, we get.

Sum of 1st and 17th terms of an arithmetic sequence is 40.

⇒ T₁ + T₁₇ = 40.

⇒ [a + (1 - 1)d] + [a + (17 - 1)d] = 40.

⇒ a + [a + 16d] = 40.

⇒ a + a + 16d = 40.

⇒ 2a + 16d = 40. - - - - - (1).

Sum of it's 1st and 18th terms of an arithmetic sequence is 43.

⇒ T₁ + T₁₈ = 43.

⇒ [a + (1 - 1)d] + [a + (18 - 1)d] = 43.

⇒ a + [a + 17d] = 43.

⇒ a + a + 17d = 43.

⇒ 2a + 17d = 43. - - - - - (2).

From equation (1) and equation (2), we get.

Subtracting both the equations, we get.

⇒ 2a + 16d = 40. - - - - - (1).

⇒ 2a + 17d = 43. - - - - - (2).

⇒ -     -          -

We get,

⇒ 16d - 17d = 40 - 43.

⇒ - d = - 3.

⇒ d = 3.

Put the value of d = 3 in equation (1), we get.

⇒ 2a + 16d = 40. - - - - - (1).

⇒ 2a + 16(3) = 40.

⇒ 2a + 48 = 40.

⇒ 2a = 40 - 48.

⇒ 2a = - 8.

⇒ a = - 4.

First term = a = - 4.

Common difference = d = 3.

To find :

(1) What is it's common difference.

Common difference : d = 3.

(2) What is the sum of it's 7th and 11th terms.

⇒ T₇ + T₁₁ = [a + (7 - 1)d] + [a + (11 - 1)d].

⇒ T₇ + T₁₁ = [a + 6d] + [a + 10d].

⇒ T₇ + T₁₁ = a + 6d + a + 10d.

⇒ T₇ + T₁₁ = 2a + 16d.

Put the values in the equation, we get.

⇒ T₇ + T₁₁ = 2(-4) + 16(3).

⇒ T₇ + T₁₁ = - 8 + 48.

⇒ T₇ + T₁₁ = 40.

Sum of it's 7th and 11th terms : T₇ + T₁₁ = 40.

(3) Find it's 9th terms.

⇒ T₉ = a + (9 - 1)d.

⇒ T₉ = a + 8d.

Put the values in the equation, we get.

⇒ T₉ = - 4 + 8(3).

⇒ T₉ = - 4 + 24.

⇒ T₉ = 20.

it's 9th terms : T₉ = 20.