Miscellaneous Exercise Class 11 Chapter 9 Sequences And Series

Let \( a \) and \( d \) be the first term and the common difference of the A.P. respectively. It is known that the kth term of an A. P. is given by
\(a_{k}={a}+({k}-1) {d}\)
\(\therefore {a}_{{m}+{n}}={a}+({m}+{n}-1) {d}\)
\({a}_{{m}-{n}}={a}+({m}-{n}-1) {d}\)
\({a}_{{m}}={a}+({m}-1) {d}\)
\(\therefore {a}_{{m}+{n}}+{a}_{{m}-{n}}={a}+({m}+{n}-1) {d}+{a}+({m}-{n}-1) {d}\)
\(=2 {a}+({m}+{n}-1+{m}-{n}-1) {d}\)
\(=2 {a}+(2 {m}-2) {d}\)
\(=2 {a}+2({m}-1) {d}\)
\(=2[{a}+({m}-1) {d}]\)
\(=2 {am}\)
Thus, the sum of \( (m+n)^{\text {th}} \) and \( (m-n)^{\text {th}} \) terms of an A.P. is equal to twice the \( m^{\text {th}} \) term.

Let the three numbers in A.P. be \( {a}-{d} \), a , and \( {a}+{d} \).
According to the given information,
\((a-d)+(a)+(a+d)=24 \ldots(1)\)
\(\Rightarrow 3 a=24\)
\(\therefore a=8\)
\((a-d) a(a+d)=440 \ldots(2)\)
\(\Rightarrow(8-d)(8)(8+d)=440\)
\(\Rightarrow(8-d)(8+d)=55\)
\(\Rightarrow 64-d^{2}=55\)
\(\Rightarrow d^{2}=64-55=9\)
\(\Rightarrow d= \pm 3\)
Therefore, when \( d=3 \), the numbers are 5, 8, and 11 and when \( d=-3 \), the numbers are 11, 8, and 5. 584
Thus, the three numbers are 5, 8, and 11.

Let a and b be the first term and the common difference of the A.P. respectively. Therefore,
\(S_{1}=\frac{n}{2}[2 a+(n-1) d] \ldots(1)\)
\(S_{2}=\frac{2 n}{2}[2 a+(2 n-1) d]=n[2 a+(2 n-1) d] \ldots(2)\)
\(S_{3}=\frac{3 n}{2}[2 a+(3 n-1) d] \ldots(3)\)
From (1) and (2), we obtain
\(\left.\left.{S}_{2}-{S}_{1}={n}[2 {a}++(2 {n}-1) {d})\right]-\frac{n}{2}[2 {a}+({n}-1) {d})\right]\)
\(={n}\left\{\frac{4 a+4 n d-2 d-2 a-n d+d}{2}\right\}\)
\(={n}\left\{\frac{2 a+3 n d-d}{2}\right\}\)
\(=\frac{n}{2}[\left(2 a+93 n-1) d\right]\)
\(\therefore 3\left({S}_{2}-{S}_{1}\right)=\frac{3 n}{2}[2 {a}+(3 {n}-1) {d}]={S}_{3}\quad[\text {from }(3)]\)
Hence, the given result is proved.

The numbers lying between \(200 \) and \(400\), which are divisible by \(7 ,\) are \(203, 210, 217 \ldots 399\)
\( \therefore \) First term, \( {a}=203 \)
Last term, \( 1=399 \)
Common difference, \( {d}=7 \)
Let the number of terms of the A.P. be \( n \).
\(\therefore {a}_{{n}}=399={a}+({n}-1) {d}\)
\(\Rightarrow 399=203+({n}-1) 7\)
\(\Rightarrow 7({n}-1)=196\)
\(\Rightarrow {n}-1=28\)
\(\Rightarrow {n}=29\)
\({S}_{29}=\frac{29}{2}(203+399)\)
\(=\frac{29}{2}(602)\)
\(=(29)(602)\)
\(=8729\)
Thus, the required sum is \(8729 \).

The integers from 1 to 100, which are divisible by 2, are \( 2,4,6 \ldots 100 \).
This forms an A.P. with both the first term and common difference equal to 2.
\(\Rightarrow 100=2+({n}-1)^{2}\)
\(\Rightarrow {n}=50\)
\(\therefore 2+4+6+\ldots+100=\frac{50}{2}[2(2)+(50-1)(2)]\)
\(=\frac{50}{2}[4+98]\)
\(=(25)(102)\)
\(=2550\)
The integers from 1 to 100, which are divisible by 5, are \( 5,10 \ldots 100 \).
This forms an A.P. with both the first term and common difference equal to 5.
\(\therefore 100=5+({n}-1)^{5}\)
\(\Rightarrow 5 {n}=100\)
\(\Rightarrow {n}=20\)
\(\therefore 5+10+\ldots+100=\frac{20}{2}[2(5)+(20-1) 5]\)
\(=10[10+(19) 5]\)
\(=10[10+95]=10 \times 105\)
\(=1050\)
The integers, which are divisible by both 2 and 5, are \( 10,20, \ldots 100 \).
This also forms an A.P. with both the first term and common difference equal to 10.
\(\therefore 100=10+({n}-1)(10)\)
\(\Rightarrow 100=10 {n}\)
\(\Rightarrow {n}=10\)
\(\therefore 10+20+\ldots+100=\frac{10}{2}[2(10)+(10-1)(10)]\)
\(=5[20+90]=5[110]=550\)
Required sum \( =2550+1050-550=3050 \)
Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

e two-digit numbers, which when divided by 4, yield 1 as remainder, are \( 13,17, \ldots 97 \).
This series forms an A.P. with first term 13 and common difference 4.
Let \( n \) be the number of terms of the A.P.
It is known that the nth term of an A.P. is given by, \( a n=a+(n-1) d \)
\(\therefore 97=13+({n}-1)(4)\)
\(\Rightarrow 4({n}-1)=84\)
\(\Rightarrow {n}-1=21\)
\(\Rightarrow {n}=22\)
Sum of \( n \) terms of an A.P. is given by,
\({S}_{{n}}=\frac{n}{2}[2 a+(n-1) d]\)
\({S}_{22}=\frac{22}{2}[22(13)+(22-1)(4)]\)
\(=11[26+84]\)
\(=1210\)
Thus, the required sum is 1210.

It is given that,
\( f(x+y)=f(x) \times f(y) \) for all \( x, y \in N\quad \ldots\ldots\text{(1)} \)
\( {f}(1)=3 \)
Taking \( x=y=1 \) in (1),
we obtain \( {f}(1+1)={f}(2)={f}(1) {f}(1)=3 \times 3=9 \)
Similarly,
\( {f}(1+1+1)={f}(3)={f}(1+2)={f}(1) {f}(2)=3 \times 9=27 \)
\( {f}(4)={f}(1+3)={f}(1) {f}(3)=3 \times 27=81 \)
\( \therefore {f}(1), {f}(2), {f}(3), \ldots \), that is \( 3,9,27, \ldots \), forms a G.P. with both the first term and common ratio equal to \(3 \).
It is known that, \( {Sn}=\frac{a\left(r^{n}-1\right)}{r-1} \)
It is given that, \( \sum_{x=1}^{n} f(x)=120 \)
\(=120=\frac{3\left(3^{n}-1\right)}{3-1}\)
\(=120=\frac{3}{2}\left(3^{n}-1\right)\)
\(=3 n-1=80\)
\(=3 n=81=34\)
\(=n=4\)
Thus, the value of n is \(4 \).

Let the sum of \( n \) terms of the G.P. be 315.
It is known that, \( {S}_{{n}}=\frac{a\left(r^{n}-1\right)}{r-1} \)
It is given that the first term a is 5 and common ratio \( r \) is \(2 \).
\(\therefore 315=\frac{5\left(2^{n}-1\right)}{2-1}\)
\(=2 {n}-1=63\)
\(=2 {n}=64=(2)^{6}\)
\(={n}=6\)
Last term of the G.P \( =6^{\text {th}} \) term \( =ar^{6-1}=(5)(2)^{5}=(5)(32)=160 \)
Thus, the last term of the G.P. is \(160 \).

Let \( a \) and \( r \) be the first term and the common ratio of the G.P. respectively.
\(a=1 a_{3}=a r^{2}=r^{2} a^{5}=a r^{4}=r^{4}\)
\(\therefore r^{2}+r^{4}=90\)
\(\Rightarrow r^{4}+r^{2}-90=0\)
\(=r^{2}=\frac{-1 \pm \sqrt{1+360}}{2}=\frac{-1 \pm \sqrt{361}}{2}=\frac{-1 \pm 19}{2}=-10 \text { or } 9\)
\(=r \pm 3 \text { (taking real roots) }\)
Thus, the common ratio of the G.P. is \( \pm 3 \).

Let the three numbers in G.P. be \(a, ar,\) and \( ar ^{2} \)
From the given condition,
\({a}+a {r}+{ar}^{2}=56\)
\(\Rightarrow {a}\left(1+{r}+{r}^{2}\right)=56 \quad \ldots\ldots\text{(1)}\)
\({a}-1, a {r}-7, {ar}^{2}-21 \text { forms an A.P. }\)
\(\therefore(a r-7)-({a}-1)=\left({ar}^{2}-21\right)-(a r-7)\)
\(\Rightarrow a {r}-{a}-6={ar}^{2}-a {r}-14\)
\(\Rightarrow {ar}^{2}-2 {ar}+{a}=8\)
\(\Rightarrow {ar}^{2}-a {r}-{ar}+{a}=8\)
\(\Rightarrow {a}\left({r}^{2}+1-2 {r}\right)=8\)
\(\Rightarrow {a}({r}-1)^{2}=8\quad \ldots\ldots\text{(2)}\)
From (1) and (2), we get
\(\Rightarrow 7\left({r}^{2}-2 {r}+1\right)=1+{r}+{r}^{2}\)
\(\Rightarrow 7 {r}^{2}-14 {r}+7-1-{r}-{r}^{2}=0\)
\(\Rightarrow 6 {r}^{2}-15 {r}+6=0\)
\(\Rightarrow 6 {r}^{2}-12 {r}-3 {r}+6=0\)
\(\Rightarrow 6 {r}({r}-2)-3({r}-2)=0\)
\(\Rightarrow(6 {r}-3)({r}-2)=0\)
When \( {r}=2, {a}=8 \)
When
Therefore, when \( {r}=2 \), the three numbers in G.P. are 8, 16, and 32.
When, \( {r}=\frac{ 1 }{ 2 } \), the three numbers in G.P. are 32, 16, and 8.
Thus, in either case, the three required numbers are 8, 16, and 32.

Let the G.P. be \( T_{1}, T_{2}, T_{3}, T_{4} \ldots T_{2 n} \).
Number of terms \( =2 {n} \)
According to the given condition,
\({T}_{1}+{T}_{2}+{T}_{3}+\ldots+{T}_{2 {n}}=5\left[{T}_{1}+{T}_{3}+\ldots+{T}_{2 {n}-1}\right]\)
\(\Rightarrow {T}_{1}+{T}_{2}+{T}_{3}+\ldots+{T}_{2 {n}}-5\left[{T}_{1}+{T}_{3}+\ldots+{T}_{2 {n}-1}\right]=0\)
\(\Rightarrow {T}_{2}+{T}_{4}+\ldots+{T}_{2 {n}}=4\left[{T}_{1}+{T}_{3}+\ldots+{T}_{2 {n}-1}\right]\)
Let the G.P. be a, \( a r, a r^{2}, a r^{3} \ldots \)
\(\therefore \frac{\operatorname{ar}\left(r^{n}-1\right)}{r-1}=\frac{4 \times a\left(r^{n}-1\right)}{r-1}\)
\(=a {r}=4 {a}\)
\(={r}=4\)
Thus, the common ratio of the G.P. is 4.

Let the A.P. be \( a, a+d, a+2 d, a +3 d \ldots a+(n-2) d, a+(n-1) d \).
Sum of first four terms \( =a+(a+d)+(a+2 d) +(a+3 d)=4 a+6 d \)
Sum of last four terms
\(=[a+(n-4) d]+[a+(n-3) d]+[a+(n-2) d]+[a+(n-1) d]\)
\(=4 a+(4 n-10) d\)
According to the given condition,
\(4 {a}+6 {d}=56\)
\(\Rightarrow 4(11)+6 {d}=56\quad[\text {Since } {a}=11 \text { (given)}]\)
\(\Rightarrow 6 {d}=12\)
\(\Rightarrow {d}=2\)
\(\therefore 4 {a}+(4 {n}-10) {d}=112\)
\(\Rightarrow 4(11)+(4 {n}-10)^{2}=112\)
\(\Rightarrow(4 {n}-10)^{2}=68\)
\(\Rightarrow 4 {n}-10=34\)
\(\Rightarrow 4 {n}=44\)
\(\Rightarrow {n}=11\)
Thus, the number of terms of the A.P. is 11.

It is given that,
\(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}\)
\(=({a}+b {x})({b}-{cx})=({b}+{cx})({a}-{bx})\)
\(={ab}-a c x+{b}^{2} {x}-{bcx}^{2}={ab}-{b}^{2} {x}+a c x-{bcx}^{2}\)
\(=2 {b}^{2} {x}=2 {acx}\)
\(={b}^{2}={ac}\)
\(=\frac{b}{a}=\frac{c}{b} \ldots(1)\)
Also, \( \frac{b+c x}{b-c x}=\frac{c+d x}{c-d x} \)
\(=({b}+{cx})({c}-{dx})=({b}-{cx})({c}+{dx})\)
\(=b c-b d x+{c}^{2} {x}-{cdx}^{2}=b c+b d x-{c}^{2} {x}+{cdx}^{2}\)
\(=2 {c}^{2} {x}=2 {bdx}\)
\(={c}^{2}=b d\)
\(=\frac{c}{d}=\frac{b}{c} \ldots(2)\)
From (1) and (2), we obtain
\(=\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\)
Thus, \( a, b, c \), and dare in G.P.

Let, \( t \) and \( d \) be the first term and the common difference of the A.P. respectively.
The \(n^{th}\) term of an A.P. is given by, \( {an}={t}+({n}-1) {d} \)
Therefore,
\(=a_{{p}}={t}+({p}-1) {d}={a} \ldots(1)\)
\(=a_{{q}}={t}+({p}-1) {d}={b} \ldots(2)\)
\(=a_{{r}}={t}+({r}-1) {d}={c} \ldots(3)\)
Subtracting eq. (2) from (1), we obtain
\(({p}-1-{q}+1) {d}={a}-{b}\)
\(=({p}-{q}) {d}={a}-{b}\)
\(={d}=\frac{a-b}{p-q} \ldots (4)\)
Subtracting eq. (3) from (2), we obtain
\(({q}-1-{r}+1) {d}={b}-{c}\)
\(=({q}-{r}) {d}={b}-{c}\)
\(={d}=\frac{b-c}{q-r} \ldots (5)\)
Equating both the values of d obtained in (4) and (5), we obtain
\(=\frac{a-b}{p-q}=\frac{b-c}{q-r}\)
\(=({a}-{b})({q}-{r})=({b}-{c})({p}-{q})\)
\(=a q-a r-b q+b r=b p-b q-c p+c q\)
\(=b p-c p+c q-a q+a r-b r=0\)
\(=(-a q+a r)+(b p-b r)+(-c p+c q)=0 \text { [by rearranging terms] }\)
\(=-a(q-r)-b(r-p)-c(p-q)=0\)
\(=a(q-r)+b(r-p)+c(p-q)=0\)
Thus, the given result is proved.

It is given that \( {a}\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right) \) are in A.P.
\({b}\left(\frac{1}{c}+\frac{1}{a}\right)-a\left(\frac{1}{b}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)-b\left(\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{b(a+c)}{a c}-\frac{a(b+c)}{b c}=\frac{c(a+b)}{a b}-\frac{b(a+c)}{a c}\)
\(=\frac{b^{2} a+b^{2} c-a^{2} b-a^{2} c}{a b c}=\frac{c^{2} a+c^{2} b-b^{2} a-b^{2} c}{a b c}\)
\(={b}^{2} {a}-{a}^{2} {b}+{b}^{2} {c}-{a}^{2} {c}={c}^{2} {a}-{b}^{2} {a}+{c}^{2} {b}-{b}^{2} {c}\)
\(={ab}({b}-{a})+{c}\left({b}^{2}-{a}^{2}\right)={a}\left({c}^{2}-{b}^{2}\right)+b c({c}-{b})\)
\(={ab}({b}-{a})+{c}({b}-{a})({b}+{a})={a}({c}-{b})({c}+{b})+b c({c}-{b})\)
\(=({b}-{a})({ab}+c b+{ca})=({c}-{b})({ac}+{ab}+b c)\)
\(={b}-{a}={c}-{b}\)
Thus, \( a, b \), and \( c \), are in A.P.

It is given that \( {a}, {b}, {c} \) and \(d\) are in G.P.
\({b}^{2}={ac} \ldots(1)\)
\({c}^{2}=b d \ldots(2)\)
\({ad}=b c \ldots(3)\)
It has to be proved that \( \left({a}^{{n}}+b^{{n}}\right),\left(b^{{n}}+c^{{n}}\right),\left(c^{{n}}+d^{{n}}\right) \) are in G.P. i.e.,
\(=\left(b^{{n}}+c^{{n}}\right)^{2}=\left({an}+b^{{n}}\right)\left(c^{{n}}+d^{{n}}\right)\)
Consider L.H.S.
\(=\left(b^{{n}}+c^{{n}}\right)^{2}={b}^{2} {n}+2 {b}^{2} {cn}+{c}^{2} {n}\)
\(=({ac})^{{n}}+2 {b}^{{n}} {c}^{{n}}+(b d)^{{n}}\quad[\text {using (1) and (2)}] \)
\(={a}^{{n}} c^{{n}}+b^{{n}} c^{{n}}+b^{{n}} c^{{n}}+b^{{n}} d^{{n}}\)
\(={a}^{{n}} c^{{n}}+b^{{n}} c^{{n}}+{a}^{{n}} d^{{n}}+b^{{n}} d^{{n}}\quad[\text {using (3)}]\)
\(=c^{{n}}\left({a}^{{n}}+b^{{n}}\right)+d^{{n}}\left({a}^{{n}}+b^{{n}}\right)\)
\(=\left({a}^{{n}}+b^{{n}}\right)\left(c^{{n}}+d^{{n}}\right) \text { R.H.S. }\)
Thus, \( \left({a}^{{n}}+b^{{n}}\right),\left(b^{{n}}+c^{{n}}\right) \), and \( \left(c^{{n}}+d^{{n}}\right) \) are in G.P.

It is given that \( a \) and \( b \) are the roots of \( x^{2}-3 x+p=0 \)
\( {a}+{b}=3 \) and \( {ab}={p} \quad \ldots\ldots\text{(1)} \)
Also, \( c \) and \( d \) are the roots of \( x^{2}-12 x+q=0 \)
\( {c}+{d}=12 \) and \( {cd}={q} \quad \ldots\ldots\text{(2)}\)
It is given that \( a, b, c, d \), are in G.P.
Let \( {a}={x}, {b}=x {r}, {c}={xr}^{2}, {d}={xr}^{3} \) from (1) and (2)
We obtain \( x+x r=3={x}(1+{r})=3 \)
\( ={xr}^{2}+{xr}^{3}=12 \)
\( ={xr}^{2}(1+{r})=12 \)
On dividing, we obtain
\(=\frac{x r^{2}(1+r)}{x(1+r)}=\frac{12}{3}\)
\(={r}^{2}=4\)
\(={r}= \pm 2\)
When \( {r}=2, {x}=\frac{3}{1+2}=\frac{3}{3}=1 \)
Where \( r=-2, x=\frac{3}{1-2}=\frac{3}{-1}=-3 \)
Case I
When \( r=2 \) and \( x=1, a b=x^{2} r=2, c d=x^{2} r^{5}=32 \)
\(=\frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}\)
i.e., \( (q+p):(q-p)=17: 15 \)
case II
When \( r=-2, x=-3, a d=x^{2} r=-18, c d=x^{2} r^{5}=-288 \)
\(=\frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}\)
Thus, in both the cases, we obtain \( (q+p):(q-p)=17: 15 \)

Let the two numbers be a and b.
A.M. \( =\frac{a+b}{2} \) and G.M. \( =\sqrt{a b} \)
According to the given condition,
\( \frac{a+b}{2 \sqrt{a b}}=\frac{m}{n}\)
\(= \frac{(a+b)^{2}}{4(a b)}=\frac{m^{2}}{n^{2}}\)
\(= (a+b)^{2}=\frac{4 a b m^{2}}{n^{2}}\)
\(= (a+b)=\frac{2 \sqrt{a b} m}{n} \ldots(1)\)
\(= (a-b)^{2}=\frac{4 a b m^{2}}{n^{2}}-4 a b=\frac{4 a b\left(m^{2}-n^{2}\right)}{n^{2}}\)
\(= (a-b)=\frac{2 \sqrt{a b} \sqrt{m^{2}+n^{2}}}{n} \ldots(2)\)
Adding (1) and (2), we obtain
\(2 {a}=\frac{2 \sqrt{a b}}{n}\left(m+\sqrt{m^{2}-n^{2}}\right)\)
\(={a}=\frac{\sqrt{a b}}{n}\left(m+\sqrt{m^{2}-n^{2}}\right)\)
Substituting the value of a in (1), we obtain
\(={b}=\frac{2 \sqrt{a b}}{n} m-\frac{\sqrt{a b}}{n}\left(m+\sqrt{m^{2}-n^{2}}\right)\)
\(=\frac{\sqrt{a b}}{n} m-\frac{\sqrt{a b}}{n} \sqrt{m^{2}-n^{2}}\)
\(=\frac{\sqrt{a b}}{n}\left(m-\sqrt{m^{2}-n^{2}}\right)\)
\(={a}: {b}=\frac{\frac{\sqrt{a b}}{n}\left(m+\sqrt{m^{2}+n^{2}}\right)}{\frac{\sqrt{a b}}{n}\left(m-\sqrt{m^{2}-n^{2}}\right)}=\frac{\left(m+\sqrt{m^{2}+n^{2}}\right)}{\left(m-\sqrt{m^{2}-n^{2}}\right)}\)
Thus, \( {a} ; {b}=\left({m}+\sqrt{m^{2}+n^{2}}\right):\left({m}-\sqrt{m^{2}-n^{2}}\right) \)

It is given that \( a, b, c \) is in A.P.
\( {b}-{a}={c}-{b} \quad \ldots\ldots\text{(1)}\)
It is given that \( {b}, {c}, {d} \), are in G.P.
\(=c^{2}=b d \quad \ldots\ldots\text{(2)}\)
Also, \( \frac{1}{c}, \frac{1}{d}, \frac{1}{e} \) are in A.P.
\(=\frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}\)
\(=\frac{2}{d}=\frac{1}{c}+\frac{1}{e} \quad \ldots\ldots\text{(3)}\)
It has to be proved that \( {a}, {c} , e ,\) are in G.P. i.e., \( {c}^{2}={ae} \)
From (1), we obtain
\(2 b=a+c\)
\(=b=\frac{a+c}{2}\)
From (2), we obtain
\(={d}=\frac{c^{2}}{b}\)
Subtracting this value in (3), we obtain
\(=\frac{2 b}{c^{2}}=\frac{1}{c}+\frac{1}{e}\)
\(=\frac{2(a+c)}{2 c^{2}}=\frac{1}{c}+\frac{1}{e}\)
\(=\frac{a+c}{c^{2}}=\frac{e+c}{e c}\)
\(=\frac{a+c}{c}=\frac{e+c}{e}\)
\(=({a}+{c}) {e}=({e}+{c}) {c}\)
\(={ae}+c e=e c+{c}^{2}\)
\(={c}^{2}={ae}\)
Thus, a, c, and e are in G.P.

\( 5+55+555+\ldots \)
Let, \( {Sn}=5+55+555+\ldots \) to n terms
\(=\frac{5}{9}[9+99+999+\ldots \text { to } {n} \text { terms}]\)
\(=\frac{5}{9}\left[(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots \text { to n terms}\right]\)
\(=\frac{5}{9}\left[\left(10+10^{2}+10^{3}+\ldots {n} \text { terms }\right)-(1+1+\ldots {n} \text { terms})\right]\)
\(=\frac{5}{9}\left[\frac{\left[10\left(10^{n}-1\right)\right.}{10-1}-n\right]\)
\(=\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{9}\right]\)
\(=\frac{50}{81}\left(10^{n}-1\right)-\frac{5 n}{9}\)

\( .6+.66+.666+\ldots \)
Let, \( {Sn}=0.6+0.66+0.666+\ldots \) to n terms
\( =6[0.1+0.11+0.111+\ldots \) to n terms\( ] \)
\(=\frac{6}{9}[0.9+0.99+0.999+\text { to n terms}]\)
\(=\frac{6}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^{2}}\right)+\left(1-\frac{1}{10^{3}}\right)+\cdots \text { to } n \text { terms}\right]\)
\(=\frac{2}{3}\left[(1+1+\cdots n \text { terms})-\frac{1}{10}\left(1+\frac{1}{10}+\frac{1}{10^{2}}+\cdots n\text { terms}\right)\right]\)
\(=\frac{2}{3}\left[n-\frac{1}{10}\left(\frac{1-\left(\frac{1}{10}\right)^{n}}{1-\frac{1}{10}}\right)\right]\)
\(=\frac{2}{3} n-\frac{2}{30} \times \frac{10}{9}\left(1-10^{-n}\right)\)
\(=\frac{2}{3} n-\frac{2}{27}\left(1-10^{-n}\right)\)

The given series is \( 2 \times 4+4 \times 6+6 \times 8+\ldots+{n} \) terms
\({n}^{\text {th}} \text { term }={a}_{{n}}=2 {n} \times(2 {n}+2)=4 {n}^{2}+4 {n}\)
\({a}_{20}=4(20)^{2}+4(20)=4(400)+80=1600+80=1680\)
Thus, the \( 20^{\text {th}} \) term of the series is 1680.

The given series is \( 3+7+13+21+31+\ldots \)
\(S=3+7+13+21+31+\ldots+a_{n-1}+a_{n}\)
\(S=3+7+13+21+\ldots+a_{n-2}+a_{n-1}+a_{n}\)
On subtracting both the eq., we obtain
\({S}-{S}=\left[3+\left(7+13+2131+\ldots+{a}_{{n}-1}+{a}_{{n}}\right)\right]\)\(-[(3+7+13+21+31+\ldots+{a}_{{n}-1}+{a}_{{n}})]\)
\({S}-{S}=3+\left[(7-3)+(13-7)+(21-13) +\ldots+\left({a}_{{n}}-{a}_{{n}-1}\right)\right]-{a}_{{n}}\)
\(0=3+[4+6+8+\ldots({n}-1) \text { terms}]-{a}_{{n}}\)
\({a}_{{n}}=3+[4+6+8+\ldots({n}-1) \text { terms}]\)
\({a}_{{n}}=3+\left(\frac{n-1}{2}\right)[2 \times 4+({n}-1-1) 2]\)
\(=3+\left(\frac{n-1}{2}\right)[8+({n}-2) 2]\)
\(=3+\left(\frac{n-1}{2}\right)[2 {n}+4]\)
\(=3+({n}-1)({n}-2)\)
\(=3+\left({n}^{2}+{n}-2\right)\)
\(={n}^{2}+{n}+1\)
\(\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k+\sum_{k=1}^{n} 1\)
\(=\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}+n\)
\(={n}\left[\frac{(n+1)(2 n+1)+3(n+1)+6}{6}\right]\)
\(={n}\left[\frac{2 n^{2}+3 n+1+3 n+3+6}{6}\right]\)
\(={n}\left(\frac{2 n^{2}+6 n+10}{6}\right)\)
\(=\frac{n}{3}\left(n^{2}+3 n+5\right)\)

From the given information
\({S}_{1}=\frac{n(n+1)}{2}\)
\({S}_{2}=\frac{n^{2}(n+1)^{2}}{4}\)
Here, \( {S}_{3}\left(1+8 {S}_{1}\right)=\frac{n^{2}(n+1)^{2}}{4}\left[1+\frac{8 n(n+1)}{2}\right] \)
\(=\frac{n^{2}(n+1)^{2}}{4}\left[4 n^{2}+4 n+1\right]\)
\(=\frac{n^{2}(n+1)^{2}}{4}(2 n+1)^{2}\)
\(=\frac{[n(n+1)(2 n+1)]^{2}}{4} \ldots(1)\)
Also, \( 9 S_{2}^{2}=9 \frac{[n(n+1)(2 n+1)]^{2}}{6^{2}} \)
\(=\frac{9}{36}[n(n+1)(2 n+1)]^{2}\)
\(=\frac{[n(n+1)(2 n+1)]^{2}}{4} \ldots(2)\)
Thus, from (1) and (2), we obtain \( 9 S_{2}^{2}={S}_{3}\left(1+8 {S}_{1}\right) \)

The \(n^{th}\) term of the given series is \( \frac{1^{3}+2^{3}+3^{3}+\cdots+n^{3}}{1+3+5+\cdots+(2 n-1)}=\frac{\left[\frac{n(n+1)}{2}\right]^{2}}{1+3+5+\cdots+(2 n-1)} \)
Here, \( 1,3,5, \ldots(2 {n}-1) \) is an A.P. with first term a, last term \( (2 {n}-1) \) and number of terms as \( n \)
\(=1+3+5+\ldots .+(2 {n}-1)=\frac{n}{2}[2 \times 1+({n}-1) 2]={n}^{2}\)
\({a}_{{n}}=\frac{n^{2}(n+1)^{2}}{4 n^{2}}=\frac{(n+1)^{2}}{4}=\frac{1}{4} n^{2}+\frac{1}{2} n+\frac{1}{4}\)
\({S}_{{n}}=\sum_{k=1}^{a} a_{k}=\sum_{k=1}^{n}\left(\frac{1}{4} k^{2}+\frac{1}{2} k+\frac{1}{4}\right)\)
\(=\frac{1}{4} \frac{n(n+1)(2 n+1)}{6}+\frac{1}{2} \frac{n(n+1)}{2}+\frac{1}{4} n\)
\(=\frac{n[(n+1)(2 n+1)+6(n+1)+6]}{24}\)
\(=\frac{n\left[2 n^{2}+3 n+1+6 n+6+6\right]}{24}\)
\(=\frac{n\left(2 n^{2}+9 n+12\right)}{24}\)

\( {n}^{\text {th}} \) term of the numerator \( ={n}({n}+1)^{2}={n}^{3}+2 {n}^{2}+{n} \)
\( {n}^{\text {th}} \) term of the denominator \( ={n}^{2}({n}+1)={n}^{3}+{n}^{2} \)
\(\frac{1 \times 2^{2}+2 \times 3^{2}+\cdots+n \times(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\cdots+n^{2} \times(n+1)}=\frac{\sum_{k=1}^{n} n_{k}}{\sum_{k=1}^{n} n_{k}}=\frac{\sum_{k=1}^{n}\left(k^{3}+2 k^{2}+k\right)}{\sum_{k=1}^{n}\left(k^{3}+k^{2}\right)} \ldots(1)\)
Here, \( \sum_{k=1}^{n}\left(k^{3}+2 k^{2}+k\right) \)
\(=\frac{n^{n}(n+1)^{2}}{6}=\frac{2 n(n+1)(2 n+1)}{4}=\frac{n(n+1)}{2}\)
\(=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2}{3}(2 n+1)+1\right]\)
\(=\frac{n(n+1)}{2}\left[\frac{3 n^{2}+3 n+8 n+4+6}{6}\right]\)
\(=\frac{n(n+1)}{12}\left[3 n^{2}+11 n+10\right]\)
\(=\frac{n(n+1)}{12}\left[3 n^{2}+6 n+5 n+10\right]\)
\(=\frac{n(n+1)}{12}[3 n(n+2)+5(n+2)]\)
\(=\frac{n(n+1)(n+2)(3 n+5)}{12} \ldots(2)\)
\(\text { Also, } \sum_{k=1}^{n}\left(k^{2}+k^{2}\right)=\frac{n^{2}(n+1)^{2}}{4}+\frac{n(n+1)(2 n+1)}{6}\)
\(=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+\frac{2 n+1}{3}\right]\)
\(=\frac{n(n+1)}{2}\left[\frac{3 n^{2}+3 n+4 n+2}{6}\right]\)
\(=\frac{n(n+1)}{12}\left[3 n^{2}+7 n+2\right]\)
\(=\frac{n(n+1)}{12}\left[3 n^{2}+6 n+n+2\right]\)
\(=\frac{n(n+1)}{12}[3 n(n+2)+1(n+2)]\)
\(=\frac{n(n+1)(n+2)(3 n+1)}{12} \ldots(3)\)
From (1), (2) and (3), we obtain
\(=\frac{1 \times 2^{2}+2 \times 3^{2}+\cdots+n \times(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\cdots+n^{2} \times(n+1)}=\frac{\frac{n(n+1)(n+2)(3 n+5)}{12}}{\frac{n(n+1)(n+2)(3 n+1)}{12}}\)
\(=\frac{3 n+5}{3 n+1}\)
Hence, proved.

It is given the farmer pays \( ₹ 6000 \) in cash.
Therefore, unpaid amount \( =₹ 12000-₹ 6000=₹ 6000 \)
According to the given conditions, the interest paid annually is \( 12 \% \) of \( 6000,12 \% \) of \( 5500,12 \% \) of \( 5000 \ldots 12 \% \) of 500
Thus, total interest to be paid
\(=12 \% \text { of } 6000+12 \% \text { of } 5500+12 \% \text { of } 5000+\ldots+12 \% \text { of } 500\)
\(=12 \% \text { of }(6000+5500+5000+\ldots+500)\)
\(=12 \% \text { of }(500+1000+1500+\ldots+6000)\)
Now, the series \( 500,1000,1500,6000 \) is an A.P. with both the first term and common difference equal to \(500\)
Let, the number of terms of the A.P. be \(n\).
\(6000=500+({n}-1) 500\)
\(=1+({n}-1)=12\)
\(={n}=12\)
Sum of A.P.
\(=\frac{12}{2}[2(500)+(12-1)(500)]=6[1000+5500]=6[6500]=39000\)
Thus total interest to be paid
\(=12 \% \text { of }(500+1000+1500+\ldots+6000))\)
\(=12 \% \text { of } 39000=₹ 4680\)
Thus, cost of tractor \( =(₹ 12000+₹ 4680)=₹ 16680 \)

It is given that sham shed Ali buys a scooter for ₹ 22000 and pays ₹ 4000 in cash.
Unpaid amount \( =₹ 22000-₹ 4000=₹ 18000 \)
According to the given condition, the interest paid annually is \( 10 \% \) of \( 18000,10 \% \) of \( 17000,10 \% \) of \( 16000 \ldots .10 \% \) of 1000 .
Thus, total interest to be paid
\(=10 \% \text { of } 18000,10 \% \text { of } 17000,10 \% \text { of } 16000 \ldots .10 \% \text { of } 1000\)
\(=10 \% \text { of }(18000+17000+16000+\ldots+1000)\)
\(=10 \% \text { of }(1000+200+3000+\ldots+18000)\)
Here, \( 1000,2000,3000 \ldots 18000 \) forms an A.P. with first term and common difference both equal to \(1000\).
Let, the numbers of terms be \( n \).
\(=18000=1000+({n}-1)(1000)\)
\(={n}=18\)
\(1000+2000+\ldots+18000=\frac{18}{2}[2(1000)+(18-1)(1000)]\)
\(=9[2000+17000]\)
\(=171000\)
Total interest paid \( =10 \% \) of \( (18000+17000+16000+\ldots+1000) \)
\( =10 \% \) of ₹ \( 171000= \) ₹ \(17100\)
Cost of scooter \( =₹ 22000+₹ 17100=₹ 39100 \)

The numbers of letters mailed forms a G.P. \( 4,42, \ldots 48 \)
First term \( =4 \)
Common ratio \( =4 \)
Numbers of item \( =8 \)
It is known that the sum of n terms of a G.P. is given by
\( {S}_{{n}}=\frac{a\left(r^{n}-1\right)}{1-r} \)
\( {S}_{8}=\frac{4\left(4^{8}-1\right)}{4-1}=\frac{4(65536-1)}{3}=\frac{4(65535)}{3}=4(21845)=87380 \)
It is given that the cost to mail one letter is 50 paisa.
Cost of mailing 87380 letters \( =₹ 87380 \times \frac{50}{100}=₹ 43690 \)
Thus, the amount spent when \( 8^{\text {th}} \) set of letter is mailed is ₹ \(43690 \).

It is given that the man deposited ₹ 10000 in a bank at the rate of \( 5 \% \)
Simple interest annually \( =\frac{5}{100} \times ₹ 10000=₹ 500 \)
Interest in first year \( =10000+500+500+\ldots+500 \)
Amount in \( 15^{\text {th}} \) year \( =₹ \)
\(=₹ 10000+14 \times ₹ 500\)
\(=₹ 10000+₹ 7000\)
\(=₹ 17000\)
Amount after 20 years \( =₹ 10000+500+500+\ldots+500 \)
\(=₹ 10000+20 \times ₹ 500\)
\(=₹ 10000+₹ 10000\)
\( =₹ 20000 \)

Cost of machine \( =₹ 15625 \)
Machine depreciates by \( 20 \% \) every year.
Therefore, its value after every year is \( 80 \% \) of the original cost i.e., \( \frac{4}{5} \) of the original cost.
Value at the end of 5 years \( =15625 \times \frac{4}{5} \times \frac{4}{5} \ldots \times \frac{4}{5}=5 \times 1024=5120 \)
Thus, the value of the machine at the end of 5 years is \( ₹=5120 \).

Let, \(x\) be the numbers of days in which 150 workers finish the work.
According to the given information,
\( 150 {x}=150+146+142+\ldots({x}+8) \) terms
The series \( 150+146+142+\ldots({x}+8) \) terms is an A.P. with first term \(146\), common differences \(-4\) and number of terms as \(( x+8 )\)
\(=150 {x}=\frac{(x+8)}{2}[2(150)+({x}+8-1)(-4)]\)
\(=150 {x}=({x}+8)[150+({x}+7)(-2)]\)
\(=150 {x}=({x}+8)(150-2 {x}-14)\)
\(=150 x=(x+8)(136-2 x)\)
\(=75 x=69 x-x^{2}+544-8 x\)
\(=x^{2}+15 x-544=0\)
\(=x^{2}+32 x-17 x-544=0\)
\(=x(x+32)-17(x+32)=0\)
\(=(x+32)(x-17)=0\)
\(=x=17 \text { or } x=-32\)
However, \( x \) cannot be
\( = \) negative \( {x}=17 \)
Therefore, originally, the number of days in which the work was completed is \(17 \). Thus, required number of days \( =(17+8)=25 \)