Sequences and Series Test - 69

Let,

First term \(=a\)

Common difference \(=d\)

Now,

\(m^{\text{th}}\) term \(=t_m=a+(m-1)d\)

Since, \(T_{m}=\frac{1 }{ n}\)

\(\therefore a+(m-1) d=\frac{1}{n}\)...(1)

Now,

\(n^{\text{th}}\) term \(=t_n=a+(n-1)d\)

Since, \(T_{n}=\frac{1 }{ m}\)

\(\therefore a+(n-1) d=\frac{1 }{ m} \)...(2)

Subtracting equation (2) from (1), we get

\(( m -1) d -( n -1) d =\frac{1 }{ n} -\frac{1 }{ m} \)

\(\Rightarrow( m - n ) d =\frac{( m - n ) }{ mn} \)

\(\Rightarrow d =\frac{1 }{ mn}\)

From equation (1), we get

\(a+\frac{(m-1) }{m n}=\frac{1 }{ n} \)

\(\Rightarrow a=\frac{1 }{ n}-\frac{(m-1) }{m n }\)

\(\Rightarrow a=\frac{\{m-(m-1)\} }{ m n} \)

\(\Rightarrow a=\frac{\{m-m+1)\} }{ m n} \)

\(\Rightarrow a=\frac{1 }{ m n}\)

Now, for \(mn^{\text{th}}\) term

\(T_{m n}=\frac{1 }{ m n}+(m n-1)\frac{1}{ m n}\)

\(\Rightarrow T_{m n}=\frac{1 }{ m n}+1-\frac{1 }{ m n}\)

\(\Rightarrow T_{m n}=1\)

Let,

First term \(= a\)

Common ratio \(= r\)

Given,

\(a-a r^{2}=768\)

\(\Rightarrow a\left(1-r^{2}\right)=768\)...(1)

\(a r^{2}-a r^{6}=240\)

\(\Rightarrow a r^{2}\left(1-r^{4}\right)=240\)...(2)

Dividing equation (2) by (1), we get

\(\frac{a r^{2}\left(1-r^{4}\right)}{a\left(1-r^{2}\right)}=\frac{240 }{768}\)

\(\Rightarrow \frac{[a r^{2}\left(1-r^{2}) (1+r^2)\right)}{a\left(1-r^{2}\right)}=\frac{240 }{768}\)

\(\Rightarrow r^2(1+r^2)=\frac{5}{16} \)

\(\Rightarrow r^2+r^4=\frac{5}{16} \)

\(\Rightarrow 16r^2+16r^4=5\)

\(\Rightarrow 16r^4+16r^2-5=0\)

Using Sridharacharaya formula, we get

\(r^2=\frac{-16\pm\sqrt{(16)^2-4 \times 16 \times-5}}{2\times 16}\)

\(\Rightarrow r^2=\frac{-16\pm\sqrt{256+320}}{32}\)

\(\Rightarrow r^2=\frac{-16\pm\sqrt{576}}{32}\)

\(\Rightarrow r^2=\frac{-16\pm24}{32}\)

\(\Rightarrow r^2=\frac{-16+24}{32}\) or \(r^2=\frac{-16-24}{32}\)

\(\Rightarrow r^2=\frac{8}{32}\) or \(r^2=\frac{-40}{32}\)

\(\Rightarrow r^2=\frac{1}{4}\) or \(r^2=\frac{-5}{4}\)

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

So, \(r=\frac{1}{2}, \frac{-1}{2}\) or \(r= \frac{\sqrt{5}}{2},\frac{\sqrt{-5}}{2}\)

Considering only the real values of \(r\), from equation (1) we get

\(a(1- \frac{1}{4})=768\)

\(\Rightarrow a\times \frac{3}{4}=768\)

\(\Rightarrow a=\frac{768\times 4}{3}\)

\(\Rightarrow a=1024\)

\(\Rightarrow a=2^{10}\)

Product of the first 21 terms \(=(a^2\times r^{20})^{10} \times ar^{10}\)

\(=a^{21} \times r^{210}\)

\(=(2^{10})^{21} \times (\frac{1}{2})^{210}\)

\(=\frac{2^{210}}{2^{210}}\)

\(=1\)

Let the three numbers in A.P. be \(a-d, a\), and \(a+d\).

Given: the sum of three numbers in A.P. is 24

\( \Rightarrow( a - d )+ a +( a + d )=24 \)

\( \Rightarrow 3 a =24 \)

\( \Rightarrow a =8\)

Also, given: the product of three numbers in A.P. is 440 .

\(\Rightarrow(a-d)(a)(a+d)=440\)

Put \(a =8\)

\( \Rightarrow(8-d)(8)(8+d)=440 \)

\( \Rightarrow(8-d)(8+d)=55 \)

\( \Rightarrow 64-d^2=55 \)

\( \Rightarrow d^2=9 \)

\( \Rightarrow d= \pm 3\)

Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.
So, the three numbers are 5, 8, and 11.

Given,

a, b, c are in AP.

First term \(=t _{1}= a\)

Second term \(=t _{2}= a + d = b\)

Third term \(=t _{3}= a +2 d = c\)

Now,

\(a+c=a+(a+2 d)=2 a+2 d\)

\(\Rightarrow a+c=2(a+d)\)

\(\Rightarrow a+c=2 b\)

Squaring both sides, we get

\((a+c)^2=2b\)

\(\Rightarrow a ^{2}+ c ^{2}+2 ac =4 b ^{2}\)

\(\Rightarrow 2 ac =4 b ^{2}- a ^{2}- c ^{2}\)

\(\Rightarrow ac =\frac{4 b ^{2}- a ^{2}- c ^{2}}{2}\)

\(\Rightarrow ac =2 b ^{2}-\frac{a ^{2}+ c ^{2}}{2}\)

Squaring both sides, we get

\((ac)^{2}=\left(2 b ^{2}-\frac{\left( a ^{2}+ c ^{2}\right)}{2}\right)^{2}\)

The given AP series is \(5,12,19, \ldots, 215\)

First term \((a)=5\)

Common differenece \((d)=12-5=7\)

\(n^{\text {th }}\) term \(=a_{n}=215\)

As we know,

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

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

\(\therefore 215=5+(n-1) \times 7\)

\(\Rightarrow 210=(n-1) \times 7 \)

\(\Rightarrow n-1=30 \)

= 31

Here, \(n\) is an odd number.

So, the middle term of the given AP will be \(\left(\frac{n+1}{2}\right)^{\text {th }}\) term.

\(\therefore\) Middle term \(=\left(\frac{ n +1}{2}\right)^{\text{th}}\) term

\(=\left(\frac{31+1}{2}\right)^{\text{th} }\) term

\(=16^{\text {th}}\) term

Now,

\(a_{16}=a+15 d\)

\(=5+15 \times 7\)

\(=5+105\)

\(=110\)

So, the middle of the given AP series is \(110\).

As we know,

Standard infinite geometric series is given by: \(a, a r, a r^{2}, a r^{3}, \ldots \ldots, a r^{n-1}, \ldots\) up to infinite terms

Sum of infinite terms of Geometric Progression (GP) is given by:

\(s _{ n }=\frac{ a }{1- r }\)

Given,

\(y=x+x^{2}+x^{3}+\ldots \ldots\) up to infinite terms.

Comparing it with standard infinite GP series, we get

\(a = x\) and \(r = x\)

As we know,

Sum of infinite terms of geometric progression is given by:

\(s _{ n }=\frac{ a }{1- r }\)

\(\therefore s _{ n }=\left(\frac{ x }{1- x }\right) \)

\(\Rightarrow y =\left(\frac{ x }{1- x }\right) \)

\(\Rightarrow y (1- x )= x \)

\(\Rightarrow y - xy = x \)

\(\Rightarrow y = x + xy \)

\(\Rightarrow y = x (1+ y ) \)

\(\Rightarrow x=\left(\frac{y}{1+y}\right)\)

Given that if middle term in the given GP is doubled then the new numbers will be in AP.

Let the three terms in GP be \(a, a r, a r^{2}\).

The middle term \(=a r\).

According to the statement given in the question,

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

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

\(\Rightarrow 4 r=1+r^{2} \)

\(\Rightarrow r^{2}-4 r+1=0\)...(1)

The above equation is a quadratic equation in terms of \(r\).

Comparing the equation (1) with \(a x^{2}+b x+c=0\), we get

\(a=1, b=-4, c=1\)

The solution of a quadratic equation is given by,

\( x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

Therefore substituting the values of \(a, b, c\) in above equation will give the value of \(r\) (common ratio).

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

\(\Rightarrow r=\frac{4 \pm \sqrt{16-4}}{2} \)

\(\Rightarrow r=\frac{4 \pm 2 \sqrt{3}}{2} \)

\(\Rightarrow r=2 \pm \sqrt{3} \)

\(\Rightarrow r=2+\sqrt{3}\)

The value of \(r\) is \(2+\sqrt{3}\), the negative value is neglected because the GP is increasing and \(2-\sqrt{3}<1\).

Let us consider two AP's having first term \(a _{1}\) and \(a _{2}\) and the common difference are \(d _{1}\) and \(d _{2}\) respectively.

According to the question,

\(\frac{ S _{1}}{ S _{2}}=\frac{3 n +8}{7 n +15} \)

As we know,

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

\(\therefore \frac{\left.\frac{ n }{2}\left[2 a _{1}+( n -1) d _{1}\right)\right]}{\left.\frac{ n }{2}\left[2 a _{2}+( n -1) d _{2}\right)\right]}=\frac{3 n +8}{7 n +15} \)

\(\Rightarrow \frac{\left.\left[ a _{1}+\left(\frac{ n -1}{2}\right) d _{1}\right)\right]}{\left[ a _{2}+\left(\frac{ n -1}{2}\right) d _{2}\right]}=\frac{3 n +8}{7 n +15} \)...(1)

As we know,

\(n^{\text{th}}\) term of AP \((a_n)=a+(n-1) \times d\)

So the ratio of \(12^{\text {th }}\) term of first AP and second AP \(=\frac{ a _{1}+(12-1) d _{1}}{ a _{2}+(12-1) d _{2}}\)

\(=\frac{ a _{1}+11 d _{1}}{ a _{2}+11 d _{2}}\)...(2)

From (1) and (2), we can write

\(\frac{ n -1}{2}=11\)

\(\Rightarrow n-1=22\)

\(\Rightarrow n=23\)

Putting \(n = 23\) in equation (1), we get

\(\frac{\left.\left[ a _{1}+\left(\frac{23 -1}{2}\right) d _{1}\right)\right]}{\left[ a _{2}+\left(\frac{ 23 -1}{2}\right) d _{2}\right]}=\frac{3 \times 23 +8}{7 \times 23 +15} \)

\(\Rightarrow \frac{ a _{1}+11 d _{1}}{ a _{2}+11 d _{2}}=\frac{69+8}{161+15}\)

\(\Rightarrow \frac{ a _{1}+11 d _{1}}{ a _{2}+11 d _{2}}=\frac{77}{176}\)

\(\Rightarrow \frac{ a _{1}+11 d _{1}}{ a _{2}+11 d _{2}}=\frac{7}{16}\)

\(\Rightarrow \frac{12^{\text {th }} \text { term of first AP}}{12^{\text {th }} \text {term of second AP}}=\frac{7}{16}\)

So, the ratio of \(12^{\text {th }}\) term of first AP and second AP is \(7:16\).

As we know,

Arithmetic mean \((A)=\frac{a+b}{2}\)

Geometric mean \((G)=\sqrt{a b}\)

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

Given,

The arithmetic mean of two numbers \(=13\)

\(\therefore \frac{a+b}{2}=13\)

\(\Rightarrow a+b=26\)...(1)

The geometric mean of two numbers \(=12\)

\(\therefore \sqrt{a b}=12\)

Squaring both sides, we get

\( a b=144 \)

\(\Rightarrow b=\frac{144 }{ a}\)

Putting the value of \(b\) in equation (1), we get

\(a+\frac{144 }{ a}=26 \)

\(\Rightarrow a^{2}-26 a+144=0\)

\(\Rightarrow(a-8)(a-18)=0 \)

\(\therefore a=8\) or \(18\)

So, \(b=26-8=18\) or

\(b= 26-18=8\)

Therefore, the number are \(8\) and \(18\).

As we know,

The general term of an geometric progression with first term \(a\) and common ratio \(r\) is given by:

\(T _{ n }= ar ^{ n -1}\)

Given,

The third term of a geometric progression (GP) \(=2\)

So, Third term \(=T _{3}= ar ^{3-1}\)

\(\Rightarrow ar ^{2}=2\)

Now, the product of first five terms \(=a \times ar \times ar^2 \times ar^3 \times ar^4\)

\(=a ^2r^{10}\)

\(=(ar^2)^5\)

\(=(2)^5\)

\(=32\)

So, the product of the first 5 terms is \(32\).