Question

Find the 8th term from the end of the sequence 3, 6, 12, ... 25th term.​

Answer 1

Answer:

24 is your answer

Step-by-step explanation:

HOPE I WILL HELP YOU

Answer 2

The 8th term from the end of the sequence 3, 6, 12, ... 25th term is 393216.

Step-by-step explanation:

Given:

The sequence 3, 6, 12, ... 25th term.

Formula Used:

For Geometric Progression

$t_{n} =pq^{n-1}$

Where,

p = First term

q= common ratio

$t_{n}$= nth term

To Find:

The 8th term from the end of the sequence 3, 6, 12, ... 25th term.

Solution:

As  given,the sequence 3, 6, 12, ... 25th term.

The sequence 3, 6, 12, ... 25th term is in Geometric Progression(G.P.).

p=3      and q $=\frac{6}{3} =2$

Applying formula no.01.

25th term  of G.P.,$t_{25} =3\times 2^{25-1}$

                                    $=3\times 2^{24}$

 We have,   3, 6, 12, ... $(3\times 2^{24})$.                                

Now, reverse the given G.P.  $(3\times 2^{24})$.........12,6,3.

Since it can be clearly observed that in the G.P   $(3\times 2^{24})$.........12,6,3. each succeeding term is obtained by dividing the preceding term by 2.  

So, the given G.P can be written as   $(3\times 2^{24})$.........12,6,3.

Hence,  p= $(3\times 2^{24})$

              q $=\frac{1}{2}$            

∴ 8th term  of G.P  $t_{8} =(3\times 2^{24})\times (\frac{1}{2} )^{8-1}$

                                     $=(3\times 2^{24})\times (\frac{1}{2} )^{7}$

                                      $=(3\times 2^{24})\times (\frac{1}{2^{7} } )$

                                      $=(3\times 2^{17})$

                                     $=393216$

Thus, the 8th term from the end of the sequence 3, 6, 12, ... 25th term is 393216.

#SPJ3