Question

What term of ap 3,10,17,........ will be 84 more than its 13th term?

Answer 1

Let nth term is 84 more than 13th term
First term = 3 and common difference = 10-3 = 7
$tn = t13 + 84 \\ = 3 + (13 - 1) \times 7 + 84 = 171 \\ so \: \: 3 + (n - 1) \times 7 = 171 \\ n = \frac{171 - 3}{7} + 1 = 25$
Thus the required term is 25th

Answer 2

Heya!!

Thanks for asking the question!

Here is your answer : -

Given,

A.P. => 3,10,17.........

To find ,

Which term of this AP will be 84 more than the 13th term of this AP.

Main Part :

a ( First term ) = 3

d ( Common difference ) = 10-7 = 17-10 = 7

n = 13

nth a = a + (n-1)d

Substituting the values on RHS , we get -

= 3 + ( 13-1 ) 7

= 3 + 12.7

= 3 + 84

= 87. ( i )

Now we know the required term of this AP is 84 more than ( i ).

So it will be -

= 87 + 84

= 171

Now we know,

d = 7

a = 3

nth term = 171

To find ,
n

Now we know,

nth term = a + ( n - 1 ) d

Substituting the values on LHS and RHS , we get -

171 = 3 + ( n - 1 )7

Now this is a Linear Equation in one Variable.

Now we will solve it for n using transposition.

The steps are as follows -

Firstly opening the brackets on RHS

171 = 3+ 7n - 7

3 + 7n - 7 = 171

7n - 4 = 171

7n = 175

n = 25

The required answer is 25th term.


Hope it helps you : )