Sequences and Series Test - 68

As we know,

If the first term of a GP is \(a\) and the common ratio is \(r\), then in this case

GP \(=a, a r, a r^{2} \ldots \ldots \)

Given that every term is equal to the sum of next two terms.

So, \(a = ar + ar ^{2}\)

Taking \(a=1\), we get

\(1=r+r^{2}\)

\(\Rightarrow r^{2}+r-1=0\)

Using Sridharacharya formula \(\left(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\right)\), to solve above equation.

\(a=1 , b=1 ,c=-1\)

\(\therefore r=\frac{-1\pm\sqrt{1^2-4\times 1\times -1}}{2\times 1} \)

\(\Rightarrow r=\frac{-1\pm\sqrt{1+4}}{2}\)

\(\Rightarrow r=\frac{-1\pm\sqrt{5}}{2}\)

\(\Rightarrow r =\frac{\sqrt{5}-1}{2}\)

Above equation can also be written as:

\(r =2 \times \frac{\sqrt{5}-1}{4}\)

As we know,

\(\sin 18^\circ =\frac{\sqrt{5}-1}{4}\)

\(\therefore r=2 \sin 18^{\circ}\)

So, the common ratio of the GP is \(2 \sin 18^{\circ}\).

As we know,

If \(x\) and \(y\) be the two numbers, then arithmetic mean \((A)\), Geometric mean \((G)\) and Harmonic mean \((H)\) of \(x\) and \(y\) is given by,

\(A=\frac{x+y}{2} \)...(1)

or \(2A= (x+y)\)...(2)

\(G^{2}=x y\)...(3)

\(H= \frac{2 x y}{x+y}\)...(4)

Given,

The harmonic mean of two number \(=4\)

\(\therefore \frac{2 x y}{x+y}=4 \)

\(\Rightarrow 2 x y=4(x+y) \)

\(\Rightarrow x y=2(x+y) \)

From equation (2) and (3), we get

\(G^{2}=2 \times 2A \)

\(\Rightarrow G^{2}=4 A \)...(5)

Arithmetic mean \((A)\) and Geometric mean \((G)\) satisfy the relation:

\(2 A+G^{2}=27 \)

\(\therefore 2 A+4A=27 \)

\(\Rightarrow 6 A=27 \)

\(\Rightarrow A=\frac{27}{6}\)

\(\Rightarrow A=\frac{9}{2}\)

From equation (1), (3) and (4), we have

\(x+y=9\) and \(x y=18\)

\(\therefore x=6 \) and \(y=3\)

So, the two numbers are \(6\) and \(3\) that satisfy the relation \(2 A + G ^{2}=27\).

For the given sequence \(-8,-5,-2, \ldots, 7\), we have

\(a=-8, d=3\) and

Last term \(\left( a_{n}\right)=7\)

\( a_{n}=a_{1}+(n-1) d \)

\(\Rightarrow 7 =-8+(n-1) 3 \)

\(\Rightarrow 7+8 =(n-1)^{3} \)

\(\Rightarrow \frac{15}{3} =(n-1) \)

\(\Rightarrow 5 =n-1 \)

\(\Rightarrow n =5+1\)

\(\Rightarrow n =6\)

As we know,

For any arithmetic sequence \(a_{1}, a_{2}, \ldots, a_{n}\), the sum of the series, \(S_{n}\left(a_{1}+a_{2}+\ldots+a_{n}\right)\) is given by,

\(S_{n}=\frac{n}{2} \times[2 \times a+(n-1) \times d]\)

Where \(a=\) First term of the sequence, \(d=\) Common difference, \(n=\) Number of terms in the sequence.

\(\therefore S_{6}=\frac{6}{2} \times[2 \times(-8)+(6-1) \times 3]\)

\(\Rightarrow S_{6}=3 \times[-16+5 \times 3]\)

\(\Rightarrow S_{6}=3 \times[-16+15]\)

\(\Rightarrow S_{6}=3 \times-1\)

\(\Rightarrow S_{6}=-3\)

So,the sum of the sequence\(-8,-5,-2, \ldots, 7\) is \(-3\).

Let \(r\) be the common ratio of the GP.

Given,

First term of the GP \(=a=1\)

Third term \(=ar^{3-1}=a r^{2} \)

Fifth term \(=ar^{5-1}=a r^{4}\)

Given,

The sum of the third term and fifth term \(=90\)

\(\therefore a r^{2}+a r^{4}=90 \)

\(\Rightarrow a\left(r^{2}+r^{4}\right)=90 \)

\(\Rightarrow 1 \times \left(r^{2}+r^{4}\right)=90 \)

\(\Rightarrow r^{2}+r^{4}=90 \)

\(\Rightarrow r^{4}+r^{2}-90=0 \)

\(\Rightarrow r^{4}+ 10r^2-9r^2-90=0\)

\(\Rightarrow (r^{2}+10) (r^{2}-9)=0\)

\(\Rightarrow (r^{2}-9)=0\)

\(\Rightarrow r^2=9\)

\(\Rightarrow r=\pm3\)

So, the common ratio of the given GP is \(\pm 3\).

The given series is \(5+9+13+\ldots+49\) which is an arithmetic progression (AP) series.

So, First term \((a)=5\)

Common difference \((d)=4\)

Last term \((l)=49=n^{th}\) term

As we know,

\(n^{th}\) term of AP is given by,

\(\therefore a+(n-1) d=49 \)

\(\Rightarrow 5+(n-1)4=49 \)

\(\Rightarrow 5-4+4n=49 \)

\(\Rightarrow 1+4n=49 \)

\(\Rightarrow 4 n=48 \)

\(\Rightarrow n=12\)

The sum of the AP series is given by,

\( S _{n}=\frac{n}{2}(a+l) \)

\(\therefore S _{n}=\frac{12}{2}(5+49)\)

\(\Rightarrow S _{n}=6 \times 54\)

\(\Rightarrow S _{n}=324\)

So, the sum of the series \(5 + 9 + 13 + … + 49\) is \(324\).

Given,

\(5+55+555+\ldots \ldots\) to \(n\) terms

\(=5 \times[1+11+111+\ldots . .\) to \(n\) terms \(]\)

\(=\frac{5}{9} \times [9+99+999+\ldots\) to \(n\) terms \(]\)

\(=\frac{5}{9} \times\left[(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots\right.\) to \(n\) terms \(]\)

\(=\frac{5}{9} \times\left[\left(10+10^{2}+10^{3}+\ldots .\right.\right.\) to \(n\) terms \(\left.)-n\right]\)

So, \(a=\frac{5}{9}, r=10\)

As we know that,

\(S_{n}=a\left(\frac{1-r^{n}}{1-r}\right)\) for \(|r|<1\)

\( S_{n}=a\left(\frac{r^{n}-1}{r-1}\right) \) for \(|r|>1 \)

As common ratio \((r)>1 \), so the sum of series is given by,

\( S_{n}=a\left(\frac{r^{n}-1}{r-1}\right) \)

\(\therefore S_n =\frac{5}{9} \times\left[\frac{10 \times\left(10^{n}-1\right)}{(10-1)}-n\right] \)

\(\Rightarrow S_n=\frac{5}{9} \times \left[\frac{10 \times\left(10^{n}-1\right)}{9}-n\right]\)

\(\Rightarrow S_n=\frac{5}{9} \times \left[\frac{10^1 \times 10^{n}-10}{9}-n\right]\)

\(\Rightarrow S_n=\frac{5}{9} \times \left[\frac{(10)^{n+1}-10}{9}-n\right]\)

\(\Rightarrow S_n=\frac{5}{9} \times \left[\frac{(10)^{n+1}-10-9n}{9}\right]\)

\(\Rightarrow S_n=\frac{5}{81} \cdot \left(10^{n+1}-9 n-10\right)\)

So, the required sum is \(\frac{5}{81} \cdot\left(10^{n+1}-9 n-10\right)\).

Given summation is a GP with common ratio \(r =-\frac{1}{2}\).

For the given GP,

First term \((a) =1\)

Common ratio \((r) =-\frac{1}{2}\)

Number of terms \((n) =8\)

As we know,

\(S_{n} =\frac{a\left(r^{n}-1\right)}{r-1} \quad [r>1] \)

\(S_{n} =\frac{a\left(1-r^{n}\right)}{1-r} \quad [r<1] \)

Since \(r<1\), the sum of the G.P is given by,

\(S_{n} =\frac{a\left(1-r^{n}\right)}{1-r}\)

Sum of first eight terms \( (S_{8})=\frac{1\left(1-\left(-\frac{1}{2}\right)^{8}\right)}{1-\left(-\frac{1}{2}\right)} \)

\(\Rightarrow S_8=\frac{1\left(1-\left(\frac{1}{256}\right)\right)}{\frac{3}{2}}\)

\(\Rightarrow S_8=\frac{2\times \left(\frac{255}{256}\right)}{3} \)

\(\Rightarrow S_8=\frac{2 \times 255}{3 \times 256} \)

\(\Rightarrow S_8=\frac{85}{128}\)

Therefore, the sum of the first 8 terms of the series \(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots\) is \(\frac{85}{128}\).

Given,

\(S _{2} S _{11}= S _{ p } S _{8}\)

The sequence \(\left\{ s _{n}\right\}\) is a geometric progression.

As we know,

\(n^{\text {th }}\) term of the GP is given by:

\(a_n={ar }^{n-1}\)

So, \(S_{2}={ar}^{2-1}={ar}\)

\(\Rightarrow S_{11}={ar}^{11-1}={ar}^{10}\)

\(\Rightarrow S _{ p }= ar ^{p-1} \)

\(\Rightarrow S _{8}={ar}^{8-1}= ar ^{7}\)

Now,

\( S_{2} S _{11}= S_{ p } S _{8} \)

\(\Rightarrow ar \times ar ^{10}= ar ^{ p -1} \times ar ^{7} \)

\(\Rightarrow r ^{ p -1}= r ^{4} \)

\(\Rightarrow p -1=4 \)

\(\Rightarrow p =5\)

Let the two numbers be \(a\) and \(b\).

As we know,

\(A M=\frac{(a+b)}{2}\) and \(G M=\sqrt{a b}\)

Given,

\(GM : AM = n: m\)

\(\Rightarrow AM : GM = m : n \)

\(\Rightarrow \frac{ a + b }{2 \sqrt{ ab }}=\frac{ m }{ n } \)

\(\Rightarrow \frac{( a + b )^{2}}{4 ab }=\frac{ m ^{2}}{ n ^{2}} \)...(i)

\(\Rightarrow \frac{( a + b )^{2}-4 ab }{4 ab }=\frac{ m ^{2}- n ^{2}}{ n ^{2}} \)

\(\Rightarrow \frac{( a - b )^{2}}{4 ab }=\frac{ m ^{2}- n ^{2}}{ n ^{2}} \)...(ii)

On dividing equation (i) and (ii), we get

\(\frac{( a + b )^{2}}{( a - b )^{2}}=\frac{ m ^{2}}{ m ^{2}- n ^{2}}\)

Squaring both sides, we get

\(\frac{a+b}{a-b}=\frac{m}{\sqrt{m^{2}-n^{2}}}\)

Using componendo and dividendo both sides, we get

\( \frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\)

\(\Rightarrow \frac{a+b+a-b}{a+b-a+b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\)

\(\Rightarrow \frac{2 a}{2 b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\)

\(\Rightarrow \frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\)

\(\Rightarrow a:b=(m+\sqrt{m^{2}-n^{2}}) :(m-\sqrt{m^{2}-n^{2}})\)

So, the ratio between two numbers is \((m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})\).

Given,

\(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\) be the harmonic mean of \(a\) and \(b\).

As we know,

The harmonic mean between two numbers \(a\) and \(b\) is given by,

\(H M=\frac{2 ab }{a+b}\)

\(\therefore \frac{2 ab }{a+b}=\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} \)

\(\Rightarrow 2 a b \times\left(a^{n}+b^{n}\right)=(a+b)\left(a^{n+1}+b^{n+1}\right) \)

\(\Rightarrow 2 a^{n+1} b+2 a^{n} b^{n+1}=a^{n+2}+b^{n+2}+a^{n+1} b+2 a^{n} b^{n+1} \)

\(\Rightarrow a^{n+2}+b^{n+2}-a^{n+1} b-a^{n} b^{n+1}=0 \)

\(\Rightarrow a^{n+1}(a-b)-b^{n+1}(a-b)=0 \)

\(\Rightarrow\left(a^{n+1}-b^{n+1}\right)(a-b)=0 \)

\(\Rightarrow \left(a^{n+1}-b^{n+1}\right)=0\)

\(\Rightarrow a^{n+1}=b^{n+1}\) (this is possible only when \(n+1=0\))

\(\therefore n = -1\)