Question

Find the 31st term of the AP ,whose fifth term is 32 and 8th term is 41.​

Answer 1

Answer:

31st term of the AP is 110.

Solution:-

Let:-

The first term of AP be a.

The common difference be d.

According to the question, we have

=>>Tn = a + (n - 1)d

=>> T5 = a + (5 - 1)d

=>> 32 = a + 4d _______eq1

And:-

=>> T8 = a + 7d

=>> 41 = a + 7d __________eq2

Subtracting eq1 from eq2 we get,

=>> (a + 7d) - (a + 4d) = 41 - 32

=>> a + 7d -a -4d = 9

=>> 3d = 9

=>> d = 3

Now,using the value of d in eq1 we get

=>> 32 = a + 4 × 3

=> 32 - 12 = a = 20

Therefore:-

=>> T31 = a + (n - 1)d

=>> T31 = 20 + 30×3

=>> T31 = 20 + 90

=>> T31 = 110

Hence:-

31st term of the AP is 110.

Step-by-step explanation:

Answer 2

Given:

• a₅ = 32
• a₈ = 41

To find:

→ a₃₁ = ?

Solution:

We know that:

aₙ = a + (n - 1)d

a₅ = 32

⇒ a + (5 - 1)d = 32

⇒ a + 4d = 32 ------(Equation 1)

a₈ = 41

⇒ a + (8 - 1)d = 41

⇒ a + 7d = 41 ------ (Equation 2)

Subtracting Equation 1 from 2,

a + 7d = 41

{-} a + 4d = 32

3d = 9

⇒ d = 9 ÷ 3

⇒ d = 3

∴ Common difference = d = 3

Substitute the value of 'd' in Equation 1 to find the value of 'a'

a + 4d = 32

⇒ a + 4(3) = 32

⇒ a + 12 = 32

⇒ a = 32 - 12

⇒ a = 20

∴ First term = a = 20

For finding the 31st term,

a₃₁ = a + (31 - 1)d

⇒ a₃₁ = 20 + 30(3)

⇒ a₃₁ = 20 + 90

⇒ a₃₁ = 110

∴ 31st term = 110