Question

the 12th term of AP 9,12,15,18 ...... is?​

Answer 1

The 12th term of the given series is 42.

Given: An arithmetic series 9, 12, 15, 18, ...

To find: The 12th term of the given series

Solution:

Let's consider the first term of the given AP be $a$, the second term be $a_{2}$ and the common difference be $d$.

Given that the first term is 9.

∵ $a = 9$ and $a_{2} = 12$

∵ $d = a_{2} - a$ (Common difference is the arithmetic difference between the consecutive terms of an AP. It is same between any two consecutive terms.)

∴ $d = 12 - 9 = 3$

Now, we know that a general term of an AP can be expressed as:

∵ $a_{n} = a + (n-1)d$   (where $a$ is the first term of the AP)

To find the 12th term, we need to find $a_{12}$

⇒ $a_{12} = 9 + (12 - 1) 3$

⇒ $a_{12} = 9 + 11*3 = 9 + 33$

⇒ $a_{12} = 42$

Hence, the 12th term of the given AP is 42

Project code - #SPJ3

Answer 2

Answer:

The 12th term of the AP is 42.

Step-by-step explanation:

As per the data given in the question,

AP= 9,12,15,18 ......

a1=9, a2=12

so, difference $d=a2-a1=12-9=3$

∴12th term will be = $a1+(n-1)d= 9+(12-1)\times3$

$=9+(11\times 3)\\=9+33\\=42$

So, the 12th term of the AP is 42.

#SPJ2