Question

Find the seventh term from the end of the series:√2,2,2√2..........32.

Answer 1

Ex9.3, 5 Which term of the following sequences: 2, 2 2, 4, is 128 2, 2 2, 4, We know that an = arn 1 where an ...

Answer 2

Here, First term (a)= √2

and, Common ratio(r)=2/√2=√2

Tn=a*r^n-1

32=√2(√2)^n-1

=> 32=√2^n

=>(√2)^10=√2^n

On comparing we get n=10 terms in  total.

Now, we will flip the series  so last term will become the first term that is 32,

and the common ratio becomes √2/2=1/√2

So, Tn=a*r^n-1

=>T7=32*(1/√2)^6

=>T7=(√2)^10-6

=>T7=(√2)^4=4

So, the seventh term from the last will be 4.

Thanks!!!