Question

find the 4th term of the GP whose 5th term is 32 and 8th term is 256

Answer 1

I think this must be the ans.

Answer 2

Answer:

The 4th term is 16

Step-by-step explanation:

A geometric sequence-where every term holds a persistent ratio to its former term.

so, if the sequence is m1, m2, m3, m4......mn

then, the common ration would be r= m2/m1=m3/m2=......mn/mn-1

Given:

5th term is 32

8th term is 256

To find the nth term the formula is an=ar^n-1

Calculation:

In this question,

The 5th term given is  32

i.e a5= ar^5-1

    a5= ar^4

    32= ar^4..... equation1

The 8th term given is 256

i.e a8= ar^8-1

    a8= ar^7

    256= ar^7.....equation 2

Dividing equation 2 by equation 1 we get

          8 = r^3

Therefore, r= 2

Now, putting this in equation 1, we get

     32= ar^4

     32= a2^4

     32= a16

       a= 2

So, now the 4th term is:

     The formula as we know;

       an= ar^n-1

       a4= 2r^4-1

           = 2×2^3

           =2×8

           =16

Therefore, the 4th term is 16

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