Question

Find the 31st

term of the arithmetic progression 5, 8, 11, 14……. Using the

appropriate formula.​

Answer 1

Answer:

a=5 d=8-5 a31=?

=3

a31 = a+30d

= 5+30(3)

= 5+90

= 95

Answer 2

Concept:

The arithmetic progression is sequence in which every consecutive numbers have constant difference.

Given:

The given arithmetic progression is

5, 8, 11, 14, ....

Find:

We are ask to find 31st term of the given arithmetic progression

Solution:

By considering the given AP

First term = a = 5

Common difference = d = 8 -5 = 3

Therefore apply the formula for nth term of the AP

$T{_{n}}=a+(n-1)d$

Substitute the value of a and d for n = 31 we get

$T{_{31}}=5+(31-1)3$

$T{_{31}}=5+90$

$T{_{31}}=95$

Hence the final answer is 95.

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