Question

From an arithmetic sequence third term is 35 and 6TH Term is 67

Answer 1

Answer:

third term is 35

a+2d=35......1

6th term is 67

a+5d=67........2

form eq 1

a= 35-2d

put a in eq2

35-2d +5d=67

35+3d=67

3d=67-35 = 32

d=32/3

put d in a

a= 35-2×32/3

a=35-64/3

a= 105-64/3

a= 41/3

Answer 2

Correct Question :

• Third term of an arthemetic sequence is 34 and 6th term is 67. a) find the common difference b) Find the first term

Solution :

• 3rd term of an A.P. = 34
• 6th term of an A.P. = 67

★ General terms of an A.P.

⇒ Tₙ = a + (n - 1)d.

⇒ T₃ = 34.

⇒ T₃ = a + (3 - 1)d.

⇒ T₃ = a + 2d.

⇒ a + 2d = 34. ⇒ (1).

⇒ T₆ = 67.

⇒ T₆ = a + (6 - 1)d.

⇒ T₆ = a + 5d.

⇒ a + 5d = 67. ⇒ (2).

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

⇒ a + 2d = 34.

⇒ a + 5d = 67.

We get,

⇒ - 3d = - 33.

⇒ d = 11.

Put the value of d = 11 in equation (1), we get.

⇒ a + 2d = 34.

⇒ a + 2(11) = 34.

⇒ a = 34 - 22.

⇒ a = 12.

→ First term = a = 12.

→ Common difference = d = 11