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:

$t _{5} = 32 \\ \implies \: a + 4d = 32 \: \: \: \: \: - - - - |1| \\ and \: \: t_{8} = 41 \\ \implies \: a + 7d = 41 \: \: \: \: - - - - |2| \\ subtracting \: |1| from \: |2| \: we \: get: \\ 3d = 9 \\ d = 3 \\ so \: a \: will \: be : a + 4(3) = 32 \\ a = 32 - 12 \\ a = 20 \\ now \: 31st \: term \: is \: recquire \\ t_{31} = a + 30d = 20 + 30(3) \\ = 20 + 90 = 110$

Step-by-step explanation:

Hope it helps u buddy :)