Question

Which term of the ap:3,15,27,39...Will be 120 more its 21st form?

Answer 1

Step-by-step explanation:

From the properties of arithmetic progressions :

• $a_n = a + (n-1)d$ , where $a_n$ is the nth term, a is the first term, n is the number of terms and d is the common difference between the terms.

Here,

Common difference : 39 - 27 = 27 - 15 = 15 - 3 = 12 .

First term : 3

Let xth term be the required term.

According to the question :

= > xth term = 120 + 21st term

= > a + ( x - 1 )12 = 120 + a + ( 21 - 1 )12

= > a + ( x - 1 ) 12 = 120 + a + ( 20 )12

= > ( x - 1 )12 = 120 + 240

= > x - 1 = 10 + 20

= > x = 29

Hence, 29th term of the given AS will be 120 more than the 21st term..

Answer 2

Answer:

n = 31

Step-by-step explanation:

First term = 3

Common difference:

=> $t_{2} - t_{1}$

= 15 - 3

= 12

Hence:

21st term = a + 20d

= 3 + 240

= 243

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

=> 243 + 120 = 3 +(n - 1) × 12

=> n - 1 = $\dfrac{360}{12}$

=> n - 1 = 30

=> n = 30 + 1

=> n = 31