Ex 9.2 Class 11 Maths Ncert Solutions

The odd integers from 1 to 2001 are \( 1, 3, 5 \ldots 1999,2001 \).
This, sequence forms an A.P.
Here, first tern, \( {a}=1 \)
Common difference, \( d=2 \)
Here, \( a+(n-1) d=2001 \)
\(=1+({n}-1)(2)=2001\)
\(=2 {n}-2=2000\)
\(={n}=1001\)
\({S}_{{n}}=\frac{n}{2}[2 {a}+({n}-1) {d}]\)
\({S}_{{n}}=\frac{1001}{2}[2 \times 1+(1001-1) \times 2]\)
\(=\frac{1001}{2}[2+1000 \times 2]\)
\(=\frac{1001}{2} \times 2002\)
\(=1001 \times 1001\)
\(=1002001\)
Thus, the sum of odd numbers from 1 to 2001 is 1002001.

The natural numbers lying between 100 and 1000, which are multiples of 5, are \( 105,110, \ldots 995 \).
Here, \( {a}=105 \) and \( {d}=5 \)
\({a}+({n}-1) {d}=995\)
\(=105+({n}-1) 5=995\)
\(=({n}-1) 5=995-105=890\)
\(={n}-1=178\)
\(={n}=179\)
\({S}_{{n}}=\frac{179}{2}[2(105)+(179-1)(5)]\)
\(=\frac{179}{2}[2(105)+(178)(5)]\)
\(=(179)(105+445)\)
\(=(179)(550)\)
\(=98450\)
Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

First term \( =2 \)
Let \( d \) be the common difference of the A.P.
Therefore, the A.P. is \( 2,2+{d}, 2+2 {d}, 2+3 {d} \)
Sum of the first five terms \( =10+10 {d} \)
Sum of the next five terms \( =10+35 {d} \)
According to given condition
\(10+10 d=\frac{1}{4}(10+5 d)\)
\(=40+40 {d}=10+5 d\)
\(=30=-5 {d}\)
\(=d=-6\)
\(a_{20}=a+(20-1) d=2+(19)(-6)=2-114=-112\)
thus, the \( 20^{\text {th}} \) term of the A.P. is \(-112\)

Let, the sum of \( n \) terms of the given A.P. be \(-25\)
It is known that
\({S}_{{n}}=\frac{n}{2}[2 {a}+({n}-1) {d}]\)
Where \( {n}= \) number of terms, \( {a}= \) first term, and \( {d}= \) common difference
Here, \( a=-6 \)
\(d=\frac{11}{2}+6=\frac{-11+12}{2}=\frac{1}{2}\)
therefore, we obtain
\(-25=\frac{n}{2}\left[2 \times(-6)+({n}-1) \frac{1}{2}\right]\)
\(=-50={n}\left[-12+\frac{n}{2}-\frac{1}{2}\right]\)
\(=-50={n}\left[-\frac{25}{2}+\frac{n}{2}\right]\)
\(=-100={n}(-25+{n})\)
\(=n^{2}-5 n-20 n+100=0\)
\(=n(n-5)-20(n-5)=0\)
\(n=20 \text { or } 5\)

It is known that the general term of an A.P. is an \( ={a}+({n}-1) {d} \)
According to the given information
\(p\) th term \( =a p=a+(p-1) d=\frac{1}{q} \ldots(1) \)
\( q \) th term \( =a q={a}+({q}-1) {d}=\frac{1}{p} \ldots (2)\)
subtracting (2) from (1), we obtain
\(({p}-1) {d}-({q}-1) {d}=\frac{1}{q}-\frac{1}{p}\)
\(=({p}-1-{q}+1) {d}=\frac{p-q}{p q}\)
\(=({p}-{q}) {d}=\frac{p-q}{p q}\)
\(={d}=\frac{1}{p q}\)
Putting the value of din (1), we obtain
\({a}+({p}-1) \frac{1}{p q}=\frac{1}{q}\)
\(={a}=\frac{1}{q}-\frac{1}{q}+\frac{1}{p q}=\frac{1}{p q}\)
\({S}_{{n}}=\frac{p q}{2}[2 {a}+({pq}-1) {d}]\)
\(=\frac{p q}{2}\left[\frac{2}{p q}+({pq}-1) \frac{1}{p q}\right]\)
\(=1+\frac{1}{2}(p q-1)\)
\(=\frac{1}{2} {pq}+1-\frac{1}{2}\)
\(=\frac{1}{2} {pq}+\frac{1}{2}\)
\(=\frac{1}{2}({pq}+1)\)

Let the sum of \( n \) terms of the given A.P. be \(116 \).
\({S}_{{n}}=\frac{n}{2}[2 {a}+({n}-1) {d}]\)
Here, \( {a}=25 \) and \( {d}-22-25=-3 \)
\(S_{n}=\frac{n}{2}[2 \times 25+(n-1)(-3)]\)
\(=116=\frac{n}{2}[50-3 n+3]\)
\(=232=n(53-3 n)=53 n-3 n^{2}\)
\(=3 n^{2}-53 n+232=0\)
\(=3 n^{2}-24 n-29 n+232=0\)
\(=3 n(n-8)-29(n-8)=0\)
\(=(n-8)(3 n-29)=0\)
\(=n=8, \text { or } n=\frac{29}{3}\)
However,
\(n\) cannot be equal to \( \frac{29}{3} \)
therefore, \( {n}=8 \)
\(a_{8}=\text {last term}=a+(n-1) d=25+(8-1)(-3)\)
\(=25+(7)(-3)=25-21\)
\(=4\)
Thus, the last term of the A.P. is \(4 \).

It is given that the \({k}^{\text {th}}\) term of the A.P. is \( 5 {k}+1 \).
\({k}^{\text {th}} \text { term}=a k=a+(k-1) d\)
\(={a}+({k}-1) {d}=5 {k}+1 {a}+{kd}-{d}=5 {k}+1\)
Comparing the coefficient of \( k \), we obtain \( d=5 a-d=1 \)
\(={a}-5=1\)
\(={a}=6\)
\({S}_{{n}}=\frac{n}{2}[2 {a}+({n}-1) {d}]\)
\(=\frac{n}{2}[2(6)+({n}-1)(5)]\)
\(=\frac{n}{2}[12+5 n-5]\)
\(=\frac{n}{2}[5 n+7]\)

It is known that:
\({S}_{{n}}=\frac{n}{2}[2 {a}+({n}-1) {d}]\)
According to the given condition,
\(\frac{n}{2}[2 {a}+({n}-1) {d}]={pn}+{qn}^{2}\)
\(=\frac{n}{2}[2 {a}+{nd}-{d}]={pn}+{qn}^{2}\)
\(={na}+{n}^{2} \frac{d}{2}-{n} \frac{d}{2}={pn}+{qn}^{2}\)
Comparing the coefficient if \( {n}^{2} \) on both sides, we obtain
\(\frac{d}{2}=q\)
\(=d=2 q\)
Thus, the common difference of the A.P. is \(2 q \).

Let, \( a_{1}, a_{2} \), and \( d_{1} d_{2} \) be the first - terms and the common difference of the first and second arithmetic progression respectively.
According to the given condition,
\( \frac{\text { sum of } n \text { terns of first A.P. }}{\text { sum of } n \text { terms of second A.P. }}=\frac{5 n+4}{9 n+6} \)
\( =\frac{\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]}{\frac{n}{2}\left[2 a_{2}+(n-1) d_{2}\right]}=\frac{5 n+4}{9 n+6} \)
\( =\frac{2 a_{1}(n-1) d_{1}}{2 a_{2}+(n-1) d_{2}}=\frac{5 n+4}{9 n+6} \ldots(1) \)
Substituting \( {n}=35 \) in (1), we obtain
\(\frac{2 a_{1}+34 d_{1}}{2 a_{2}+34 d_{2}}=\frac{5(35)+4}{9(35)+5}\)
\(=\frac{a_{1}+17 d_{1}}{a_{2}+17 d_{2}}=\frac{179}{321} \ldots(2)\)
\(\frac{18^{\text {th}} \text { term of first A.P. }}{18^{\text {th}} \text { term of second A.P. }}=\frac{a_{1}+17 d_{1}}{a_{2}+17 d_{2}} \ldots (3)\)
\(\frac{18_{\text {th}} \text { term of first A.P. }}{18_{\text {th}} \text { term of second A.P. }}=\frac{179}{321}\)
Thus, the ration of \( 18^{\text {th}} \) term of both the A.P. is \( 179: 321 \).

Let, \( a \) and \( d \) be the first term and the common difference of the A.P. respectively.
Here,
\(S p=\frac{p}{2}[2 {a}+({p}-1) {d}]\)
\(S q=\frac{q}{2}[2 {a}+({q}-1) {d}]\)
According to the given condition,
\(\frac{p}{2}[2 {a}+({p}-1) {d}]=\frac{q}{2}[2 {a}+({q}-1) {d}]\)
\(={p}[2 {a}+({p}-1) {d}]={q}[2 {a}+({q}-1) {d}]\)
\(=2 {ap}+p d(p-1)=2 {aq}+q d(q-1)\)
\(=2 {a}({p}-{q})+{d}[{p}({p}-1)-{q}({q}-1)]=0\)
\(=2 {a}({p}-{q})+{d}\left[{p}_{2}-{p}-{q}^{2}+q\right]=0\)
\(=2 {a}({p}-{q})+{d}[({p}-{q})({p}+{q})-({p}-{q})]=0\)
\(=2 {a}({p}-{q})+{d}[({p}-{q})({p}+{q}-1)]=0\)
\(=2 {a}+{d}({p}+{q}-1)=0\)
\(={d}=\frac{-2 a}{p+q-1} \ldots(1)\)
\(S_{{p}+{q}}=\frac{p+q}{2}[2 {a}+({p}+{q}-1) \cdot {d}]\)
\(=S_{{p}+{q}}=\frac{p+q}{2}\left(2 a+(p+q-1)\left(\frac{-2 a}{p+q-1}\right)\right)\quad[\text {from (1)] }\)
\(=\frac{p+q}{2}[2 {a}-2 {a}]\)
\(=0\)
Thus, the sum of first \( (p+q) \) terms pf the A.P. is \(0\).

Let, \( a_{1} \) and \( d \) be the first term and the common difference of the A.P. respectively.
According to the given information,
\(S_{{p}}=\frac{p}{2}\left[2 {a}_{1}+({p}-1) {d}\right]={a}\)
\(=2 {a}_{1}+({p}-1) {d}=\frac{2 a}{p} \ldots(1)\)
\(S_{q}=\frac{q}{2}\left[2 {a}_{1}+({q}-1) {d}\right]={b}\)
\(=2 {a}_{1}+({q}-1) {d}=\frac{2 b}{q} \ldots(2)\)
\(S_{{r}}=\frac{r}{2}\left[2 {a}_{1}+({r}-1) {d}\right]={c}\)
\(=2 {a}_{1}+({r}-1) {d}=\frac{2 c}{r} \ldots(3)\)
Subtracting (2) from (1), we obtain
\(({p}-1) {d}-({q}-1) {d}=\frac{2 a}{p}-\frac{2 b}{q}\)
\(={d}({p}-1-{q}+1)=\frac{2 a q-2 b p}{p q}\)
\(={d}({p}-{q})=\frac{2 a q-2 b p}{p q}\)
\(={d}=\frac{2(a q-b p)}{p q(p-q)}\ldots(4)\)
Subtracting (3) from (2), we obtain
\(({q}-1) {d}-({r}-1) {d}=\frac{2 b}{q}-\frac{2 c}{r}\)
\(={d}({q}-1-{r}+1)=\frac{2 b}{q}-\frac{2 c}{r}\)
\(={d}({q}-{r})=\frac{2 b r-2 c q}{q r}\)
\(={d}=\frac{2(b r-q c)}{q r(q-r)} \ldots(5)\)
Equating both the value of \( d \) obtained in (4) and (5), we obtain
\(\frac{a q-b p}{p q(p-q)}=\frac{b r-q c}{q r(q-r)}\)
\(=q r(q-r) (a q-b p)=p q(p-a)(b r-q c)\)
\( =r(a q-b p)(q-r)=p(b r-q c)(p-a)\)
\(=(a q r-b p r)(q-r)=(b p r-p q c)(p-q)\)
Dividing both sides by \( p q r \), we obtain
\(\left(\frac{a}{p}-\frac{b}{q}\right)({q}-{r})=\left(\frac{b}{q}-\frac{c}{r}\right)({p}-{q})\)
\(\left(\frac{a}{p}-\frac{b}{q}\right)(q-r)=\left(\frac{b}{q}-\frac{c}{r}\right)(p-q)\)
\(=\frac{a}{p}(q-r)-\frac{b}{q}(q-r+p+q)+\frac{c}{r}(p-q)=0\)
\(=\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0\)
Thus, the given result is proved.

Let, \( a \) and \( d \) be the first term and the common difference of the A.P. respectively.
According to the given information,
\(\frac{\text { sum of } m \text { terms }}{\text { sum of } n \text { terms }}=\frac{m^{2}}{n^{2}}\)
\(=\frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) d]}=\frac{m^{2}}{n^{2}}\)
\(=\frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n} \ldots(1)\)
Putting \( {m}=2 {m}-1 \) and \( {n}=2 {n}-1 \) in (1), we obtain
\( \frac{2 a+(2 m-2) d}{2 a+(2 n-2) d}=\frac{2 m-1}{2 n-1} \)
\( =\frac{a+(m-1) d}{a+(n-1) d}=\frac{2 m-1}{2 n-1} \ldots (2)\)
\( \frac{m^{\text {th}} \text { term of A.P. }}{n_{\text {th}} \text { term of A.P. }}=\frac{a+(m-1) d}{a+(n-1) d} \ldots(3)\)
From (2) and (3), we obtain
\( \frac{m^{\text {th}} \text { term of A.P. }}{n^{\text {th}} \text { term of A.P. }}=\frac{2 m-1}{2 n-1} \)
Thus, the given result is proved.

Let, \( a \) and \( d \) be the first term and the common difference of the A.P. respectively.
\(a_{m}=a+(m-1) d=164 \ldots(1)\)
sum of n terms:
\(s_{{n}}=\frac{n}{2}[2 {a}+({n}-1) {d}]\)
here,
\(=\frac{n}{2}[2 {a}+{nd}-{d}]=3 {n}^{2}+5 {n}\)
\(=n {a}+{n}^{2} \cdot \frac{d}{2}=3 {n}^{2}+5 {n}\)
Comparing the coefficient of \( {n}^{2} \) on both sides, we obtain
\(\frac{d}{2}=3\)
\(=d=6\)
Comparing the coefficient of \( n \) on both sides, we obtain
\(={a}-\frac{d}{2}=5\)
\(={a}-3=5\)
\(={a}=8\)
Therefore, from (1), we obtain
\(8+({m}-1) 6=164\)
\(=({m}-1) 6=164-8=156\)
\(={m}-1=26\)
\(={m}=27\)
Thus, the value of \( m \) is \(27 \).

Let, \( A_{1}, A_{2}, A_{3}, A_{4} \), and \( A_{5} \) be five numbers between 8 and 26 such that \( 8, {A}_{1}, {A}_{2}, {A}_{3}, {A}_{4}, {A}_{5}, 26 \) is an A.P.
Here, \( a=8 b=26, n=7 \)
Therefore, \( 26=8+(7-1) {d} \)
\(=6 {d}=26-8=18\)
\(={d}=3\)
\({A}_{1}={a}+{d}=8+3=11\)
\({A}_{2}={a}+2 {d}=8+2 \times 3=8+6=14\)
\({A}_{3}={a}+3 {d}=8+3 \times 3=8+9=17\)
\({A}_{4}={a}+4 {d}=8+4 \times 3=8+12=20\)
\({A}_{5}={a}+5 {d}=8+5 \times 3=8+15=23\)
Thus, the required five numbers between 8 and 26 are \( 11,14,17,20 \), and \(23\).

A.M. of a and \( {b}=\frac{a+b}{2} \)
According to the given condition,
\(\frac{a+b}{2}=\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}\)
\(=(a+b)\left(a^{n-1}+b^{n-1}\right)=2\left(a^{n}+b^{n}\right)\)
\(=a^{{n}}+{a}^{{n}-1}+{ba}^{{n}-1}+b^{{n}}=2 {a}^{{n}}+2 {b}^{{n}}\)
\(={a} {b}^{{n}-1}+{a}^{{n}-1} {b}={a}^{{n}}+b^{{n}}\)
\(={ab}^{{n}-1}-b^{{n}}={a}^{{n}}-{a}^{{n}-1} {b}\)
\(={b}^{{n}-1}({a}-{b})={a}^{{n}-1}({a}-{b})\)
\(={b}^{{n}-1}={a}^{{n}-1}\)
\(=\left(\frac{a}{b}\right)^{n-1}=1=\left(\frac{a}{b}\right)^{0}\)
\(={n}-1=0\)
\(={n}=1\)

Let \( A_{1}, A_{2} \ldots A m \) be \( m \) numbers such that \( 1, A_{1}, A_{2}, \ldots A m, 31 \) is an A.P.
Here, \( {a}=1, {b}=31, {n}={m}+2 \)
\(31=1+({m}+2-1) {d}\)
\(=30=({m}+1) {d}\)
\(={d}=\frac{30}{m+1} \ldots(1)\)
\({A}_{1}={a}+{d}\)
\({A}_{2}={a}+2 {d}\)
\({A}_{3}={a}+3 {d}\)
\(\therefore {A}_{7}={a}+7 {d}\)
\({A}_{{m}-1}={a}+({m}-1) {d}\)
According to the given condition,
\(\frac{a+7 d}{a+(m-1) d}=\frac{5}{9}\)
\(=\frac{1+7\left(\frac{30}{m+1}\right)}{1+(m-1)\left(\frac{30}{m-1}\right)}=\frac{5}{9}\quad[\text {from }(1)]\)
\(=\frac{m+1+7(30)}{m+1+30(m-1)}=\frac{5}{9}\)
\(=\frac{m+1+210}{m+1+30 m-30}=\frac{5}{9}\)
\(=\frac{m+211}{31 m-29}=\frac{5}{9}\)
\(=9 {m}+1899=155 {m}-145\)
\(=155 {m}-9 {m}=1899+145\)
\(=146 {m}=2044\)
\(={m}=14\)
Thus, the value of \( m \) is \(14 \).

The first installment of the loan is ₹ 100.
The second installment of the loan is \( ₹ 105 \) and so on.
The amount that the man repays every month forms an A.P.
The A.P. is \( 100,105,110 \ldots \)
First term, \( a=100 \)
Common difference, \( d=5 \)
\(A_{30}=a+(30-1) d\)
\(=100+(29)(5)\)
\(=100+145\)
\(=245\)
Thus, the amount to be paid in the \( 30^{\text {th}} \) installment is ₹ 245.

The angles of the polygon will form an A.P. with common difference d as \( 5^{\circ} \) and first term an as \(120 \).
It is known that the sum of all angles of the polygon with \(n\) sides \( 180^{\circ} ( n -2) \)
\({S}_{{n}}=180^{\circ}({n}-2)\)
\(=\frac{n}{2}[2 {a}+({n}-1) {d}]=180^{\circ}({n}-2)\)
\(=\frac{n}{2}\left[240^{\circ}+({n}-1) 5^{\circ}\right]=180^{\circ}({n}-2)\)
\(={n}[240+({n}-1) 5]=360({n}-2)\)
\(=240 {n}+5 {n}^{2}-5 {n}=360 {n}-720\)
\(=5 {n}^{2}+235 {n}-360 {n}+720=0\)
\(=5 {n}^{2}-125 {n}+720=0\)
\(={n}^{2}-25 {n}+144=0\)
\(={n}^{2}-16 {n}-9 {n}+144=0\)
\(={n}({n}-16)-9({n}-16)=0\)
\(=({n}-9)({n}-16)=0\)
\(={n}=9 \text { or } 16\)