Question

find the 12th
term of
an A.P
$\sqrt{2}$
$3 \sqrt{2}$
$5$
$\sqrt{2}$



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Answer 1

Given AP :

√2,3√2,5√2........

To Find:

12th term of the AP.

Solution:

We can observe the common difference here as (3√2-√2)=2√2 & (5√2-3√2)=2√2 , which is constant.

Now here ,

• First term= √2 (a)
• Common Difference =2√2 (d)
• nth term = 12 (n)

So , general term of an AP is given by ,

$\large\red{\boxed{\green{\bf{\leadsto T_{n}=a+(n-1)d}}}}$

$\sf{\implies T_{12}}$=√2+(12-1)2√2.

= √2+11×2√2.

=√2+22√2

=23√2.

Answer 2

Answer:

From the given A. P. , we get:

First term=a=√2

Common difference=d=2√2

Step-by-step explanation:

From the general term of an A. P :

[Tn = {a+(n-1) d}]

==>T¹²=√2+(12-1) 2√2)

=√2+(11) 2√2

=√2+22√2=23√2 {Ans}

hope it help