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.

Answer 2

Suppose:-

• The common difference be d.
• The first term of AP be a.

Now:-

= Tn = a + (n - 1)d

= T5 = a + (5 - 1)d

= 32 = a + 4d _______(i)

Also:-

= T8 = a + 7d

= 41 = a + 7d __________(ii)

Now we need to subtract (i) from (ii) we will get,

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

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

= 3d = 9

= d = 9/3

= d = 3

By using value of the (i) here

= 32 = a + 4 × 3

= 32 - 12 = a = 20

Hence:-

= T31 = a + (n - 1)d

= T31 = 20 + 30×3

= T31 = 20 + 90

= T31 = 110

Therefore:-

31st term of the AP is 110.