Question

S is a geometric sequence? a)Given that (`sqrtx`-1), 1 and (`sqrtx`+1) are the 1st 3 terms of S, find the value of `x`. b)Show that the 5th term of S is 7+5`sqrt2`

Answer 1

a)`x=2`
b) see below

Explanation:

a) Since the first three terms are `sqrt x-1`, 1 and `sqrt x + 1`, the middle term, 1, must be the geometric mean of the other two. Hence

`1^2=(sqrt x-1)(sqrt x +1) implies`

`1= x-1 implies x=2`

b)

The common ratio is then `sqrt 2+1`, and the first term is `sqrt 2-1`.

Thus, the fifth term is

`(sqrt 2-1)times (sqrt 2+1)^4=(sqrt 2+1)^3`
`qquad = (sqrt 2)^3+3(sqrt2)^2+3(sqrt2)+1`
`qquad= 2sqrt2+6+3sqrt2+1`
`qquad =7+5sqrt2`

Answer 2

Please see below.

Explanation:

Given that,

`rarrsqrtx-1,1,sqrtx+1` are in `GP`.

So,

`rarr(sqrtx-1)/1=1/(sqrtx+1)`

`rarr(sqrtx-1)^2=1`

`rarr(sqrtx)^2-1^2=1`

`rarrx=2`

The first term `(a)=sqrtx-1=sqrt2-1`

The second term `(b)=1`

The common ratio `(r)=b/a=1/(sqrt2-1)=sqrt2+1`

The `n^(th)` term of geometric sequence `(t_n)=a*r^(n-1)`

So, `t_5=(sqrt2-1)*(sqrt2+1)^(5-1)`

`=(sqrt2-1)(sqrt2+1)(sqrt2+1)^3`

`=[(sqrt2)^2-1^2][(sqrt2)^3+3*(sqrt2^2)*1+3*sqrt2*1^2+1^3]`

`=(2-1)(2sqrt2+6+3sqrt2+1)=7+5sqrt2`

Answer 3

`x=2 and 5^(th)" term"=7+5sqrt2`.

Explanation:

For any `3` consecutive terms `a,b,c` of a GP, we have,

`b^2=ac`.

Hence, in our case, `1^2=(sqrtx-1)(sqrtx+1)=(sqrtx)^2-1^2,`

`i.e., 1=x-1, or, x=2`.

With `x=2`, the `1^(st) and 2^(nd)` terms of the GP under

reference are, `sqrtx-1=sqrt2-1 and 1`, resp.

So, the common ratio `r=(2^(nd)" term)"-:(1^(st)" term)"`,

`=1/(sqrt2-1)=sqrt2+1`.

`:. 4^(th)" term=r("3^(rd)" term)=(sqrt2+1)(sqrtx+1)`,

`=(sqrt2+1)(sqrt2+1)`,

`=2+2sqrt2+1`,

`=3+2sqrt2`.

Further, `(5^(th)" term)=r("4^(th) term)`,

`=(sqrt2+1)(3+2sqrt2)`,

`=3sqrt2+3+2sqrt2*sqrt2+2sqrt2`.

`rArr 5^(th)" term"=7+5sqrt2`.