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

Answer:

31st term = 110

Step-by-step explanation:

5th term=32

8th term=41

we have to find the common difference

32+3d=41

3d= 41 -32

3d = 9

d = 9/3

common difference = 3

so, the A P = 20,23,26,29,32,35,38,41.......

Then we have to find the 31st term

X1 is the first term

eqn, X1 +(n-1) ×d

=20+(31-1)×3

=20+30×3

=20+90

=110.

Therefore

31st term of AP = 110