Question

the 6th and 11th terms of an arithmetic sequence are 38 and 73..
(a) find the first term ​

Answer 1

Let the first term of the Arithmetic sequence be a and common difference be d

6th term of sequence = 38

a + 5d = 38 ---(1)

11th term of sequence = 73

a + 10d = 73 ---(2)

Subtract (1) from (2)

a + 10d - ( a + 5d ) = 73 - 38

a + 10d - a - 5d = 35

5d = 35

d = 35/5

d = 7

Substitute d in (1)

a + 5d = 38

a + 5(7) = 38

a + 35 = 38

a = 38 - 35

a = 3

Therefore the first term of the Arithmetic sequence is 3.

Answer 2

EXPLANATION.

6th and 11th terms of an arithmetic sequence are 38 and 73.

As we know that,

General terms of an ap.

⇒ Tₙ = a + (n - 1)d.

Using this formula in this equation, we get.

⇒ T₆ = 38.

⇒ a + (6 - 1)d = 38.

⇒ a + 5d = 38. - - - - - (1).

⇒ T₁₁ = 73.

⇒ a + (11 - 1)d = 73.

⇒ a + 10d = 73. - - - - - (2).

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

Subtracting both the equation, we get.

⇒ a + 5d = 38. - - - - - (1).

⇒ a + 10d = 73. - - - - - (2).

⇒ -  -           -

We get,

⇒ 5d - 10d = 38 - 73.

⇒ - 5d = - 35.

⇒ 5d = 35.

⇒ d = 7.

Put the value of d = 7 in the equation (1), we get.

⇒ a + 5d = 38. - - - - - (1).

⇒ a + 5(7) = 38.

⇒ a + 35 = 38.

⇒ a = 38 - 35.

⇒ a = 3.

First term = a = 3.

Common difference = d = 7.

∴ First term = a = 3.