Question

The sixth term of AP is 5 times the first term and the eleventh term exceed twice the fifth term by 3. Find eighth term?

Answer 1

ANSWER:

Let the first term be ‘a’

Given

• Sixth term( t₆ ) is 5 times the first term
• 5a = t6

Using

tn = a + (n - 1)d

• 5a = a + (6 - 1)d
• 5a = a + 5d
• 4a = 5d
• 4a - 5d = 0 ----------------( 1 )

Given

• 11th term ( t₁₁ ) exceeds twice the fifth term( t₅ ) by 3
• t₁₁ = 2(a + 4d) + 3
• a + 10d = 2a + 8d + 3
• - 3 = a - 2d

Multiplying both sides with 4

• - 12 = 4a - 8d --------------( 2 )

Subtracting Equation ( 2 ) - ( 1 )

4a - 8d = - 12

- 4a - 5d = 0

(-) (+) (-)

- 3d = - 12

→ -3d = - 12

→ d = -12/-3

→ Common difference ( d ) = 4

Substituting d = 4 in equation 2

4a - 8(4) = - 12

4a = - 12 + 32

4a = 20

a = 20/4 = 5

a = 5

TO FIND:

Eighth term ( t₈ )

Using

tn = a + (n - 1)d

t₈ = 5 + (8 - 1)4

t₈ = 5 + 28

t₈ = 33

Hence , Eighth term ( t₈ ) = 33