Question

4, 12,20,... 41st term?

Answer 1

Answer:

a= 4

a2= 12

d= a2-a

d= 12-4 = 8

now, we have to find 41 term that is a41

a41 = a+ 40d

a41= 4+ 40× 8

a41= 4+ 320

= 324

therefore the 41st term is 324

hope it helped u!!!!

Answer 2

Answer:

The 41st term of the given series is 324

Step-by-step explanation:

$\red{ \rm Given:}$

4, 12, 20,......

$\red {\rm To \: find:}$

The 41st term of the given series.

$\red{\rm How \: to \: find:}$

First we are going to check which type of series is it [AP (or) GP (or) HP)].

Then after finding the type of series we are going to use its general term expression to find the 41st term.

$\red{\rm Solution:}$

We know that AP is a series where there is a common difference between each term.

We also know that GP is a series where there is a common ratio between each term

So now let's check whether there is a common difference or common ratio between the terms.

4, 12, 20,.......

a (First term of a series) = 4

We know that d = tn - $\rm t_{n-1}$

Now let's check the difference between the second and first term and then third and second term

d = t2 - t1 = 12 - 4 = 8

d = t3 - t2 = 20 - 12 = 8

As in both the cases the difference can same therefore it is an AP

We know that the tn term of AP = a + (n - 1)d

$\boxed{\rm t_n = a + (n-1)d}$

Using the above formula,

tn = 4 + (41 - 1) * 8 = 4 + (40 * 8) = 4 + 320 = 324

$\boxed{\rm The \: 41st \: term \: of \: the \: given \: series \: is \: 324}$

$\red{\rm Formulas \: related \: to \: Arithmetic \: Progession:}$

i) General form of AP is

a, a + d, a + 2d,.......... a + (n - 1)d

ii) tn = a + (n - 1)d

iii) Sn = n[2a + (n - 1)d]/2 = n(a + l)/2, Where a is the first term and l is the last term