Sequences and Series

Class 11 NCERT

NCERT

1   Write the first five terms of the sequences whose $n^{th}$ term is $ a_n=n(n+2).$

Solution :

$a_n=n(n+2)$$\\$ Substituating $n=1,2,3,4$ and $5,$ we obtain$\\$ $ a_1=1(1+2)=3\\ a_2=2(2+2)=8\\ a_3=3(3+2)=15\\ a_4=4(4+2)=24\\ a_5=5(5+2)=35$$\\$ Therefore, the required terms are $3, 8, 15, 24$ and $35.$

2   Write the first five terms of the sequences whose $n^{th}$ term is $ a_n=\dfrac{n}{n+1}$

Solution :

$ a_n=\dfrac{n}{n+1}$$\\$ Substituting $ n=1,2,3,4,5,$ we obtain$\\$ $a_1=\dfrac{1}{1+1}=\dfrac{1}{2},a_2=\dfrac{2}{2+1}=\dfrac{2}{3},\\ a_3=\dfrac{3}{3+1}=\dfrac{3}{4},a_4=\dfrac{4}{4+1}=\dfrac{4}{5},\\ a_5=\dfrac{5}{5+1}=\dfrac{5}{6}$$\\$ Therefore, the required terms are $\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5}$ and $\dfrac{5}{6}$

3   Write the first five terms of the sequences whose $n^{th}$ term is $ a_n=2^n$

Solution :

$a_n=2^n$$\\$ Substituting $n=1,2,3,4,5,$ we obtain$\\$ $a_1=2^1=2\\ a_2=2^2=4\\ a_3=2^3=8\\ a_4=2^4=16\\ a_5=2^5=32$$\\$ Therefore, the required terms are $2, 4, 8, 16$ and $32.$

4   Write the first five terms of the sequences whose $n^{th}$ term is $ a_n=\dfrac{2n-3}{6}$

Solution :

Substituting $n = 1,2,3,4,5 ,$ we obtain$\\$ $a_1=\dfrac{2*1-3}{6}=\dfrac{-1}{6}\\ a_2=\dfrac{2*2-3}{6}=\dfrac{1}{6}\\ a_3=\dfrac{2*3-3}{6}=\dfrac{3}{6}=\dfrac{1}{2}\\ a_4=\dfrac{2*4-3}{6}=\dfrac{5}{6}\\ a_5=\dfrac{2*5-3}{6}=\dfrac{7}{6}$$\\$ Therefore, the required terms are $ \dfrac{-1}{6},\dfrac{1}{6},\dfrac{1}{2},\dfrac{5}{6}$ and $ \dfrac{7}{6}.$

5   Write the first five terms of the sequences whose $n^{th}$ term is $ a_n=(-1)^{n-1}5^{n+1}$

Solution :

Substituting $n = 1,2,3,4,5 ,$ we obtain $\\$ $a_1=(-1)^{1-1}5^{1+1}=5^2=25\\ a_2=(-1)^{2-1}5^{2+1}=-5^3=-125\\ a_3=(-1)^{3-1}5^{3+1}=5^4=625\\ a_4=(-1)^{4-1}5^{4+1}=-5^5=-3125\\ a_5=(-1)^{5-1}5^{5+1}= 5^6=15625$$\\$ Therefore, the required terms are $25, - 125,625, - 3125$ and $15625 .$

6   Write the first five terms of the sequences whose $n^{th} $ term is $ a_n=n\dfrac{n^2+5}{4}$

Solution :

Substituting $n=1,2,3,4,5,$ we obtain$\\$ $ a_1=1.\dfrac{1^2+5}{4}=\dfrac{6}{4}=\dfrac{3}{2}\\ a_2=2.\dfrac{2^2+5}{4}=2.\dfrac{9}{4}=\dfrac{9}{2}\\ a_3=3.\dfrac{3^2+5}{4}=3.\dfrac{14}{4}=\dfrac{21}{2}\\ a_4=4.\dfrac{4^2+5}{4}=21\\ a_5=5.\dfrac{5^2+5}{4}=5.\dfrac{30}{4}=\dfrac{75}{2}$$\\$ Therefore, the required terms are $ \dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21 $ and $ \dfrac{75}{2}$

7   Find the $17^{ th}$ and $24^{ th}$ term in the following sequence whose $n^{ th}$ term is $a _n = 4 _n - 3$

Solution :

Substituting $n=17,$ we obtain$\\$ $a_17=4(17)-3=68-3=65$$\\$ Substituting $ n=24, $ we obtain $\\$ $a_{24}=4(24)-3=96-3=93$

8   Find the $7^{th}$ term in the following sequence whose $n^{th}$ term is $a_ n=\dfrac{n^2}{2^n}$

Solution :

Substituting $n=7,$ we obtain$\\$ $ a_7=\dfrac{7^2}{2^7}=\dfrac{49}{128}$

9   Find the $9^{ th }$ term in the following sequence whose $n ^{th}$ term is $ a_ n=(-1)^{n-1} n^3$

Solution :

Substituting $n = 7$ , we obtain$\\$ $a_9=(-1)^{9-1}(9)^3=(9)^3=729$

10   Find the $20^{ th}$ term in the following sequence whose $n^{ th}$ term is $a_ n=\dfrac{n(n-2)}{n+3}$

Solution :

Substituting $n = 20$, we obtain $\\$ $a_{20}=\dfrac{20(20-2)}{20+3}=\dfrac{20(18)}{23} =\dfrac{360}{23}$

11   Write the first five terms of the following sequence and obtain the corresponding series: $a_ 1 = 3, a _n = 3 a_{n -1 }+ 2$ for all $n > 1$

Solution :

$a_ 1 = 3, a _n = 3 a_{n -1 }+ 2$for $n > 1$ $\\$ $\implies a_2 = 3 a_1 + 2 = 3 ( 3 )+ 2 = 11\\ a_3 = 3 a_2 + 2 = 3 ( 11 )+ 2 = 35\\ a_4 = 3 a_3 + 2 = 3 ( 35 )+ 2 = 107\\ a_5 = 3 a_4 + 2 = 3 (107 )+ 2 =323\\$ Hence, the first five terms of the sequence are 3, 11, 35, 107 and 323.$\\$ The corresponding series is $3 + 11 + 35 + 107 + 323 + ....$

12   Write the first five terms of the following sequence and obtain the corresponding series: $\\$ $a_1=-1,a_n=\dfrac{a_{n-1}}{n},n\geq 2 $

Solution :

$a_1=-1,a_n=\dfrac{a_{n-1}}{n},n\geq 2 $ $\\$ $\implies a_2=\dfrac{a_1}{2}=\dfrac{-1}{2}\\ a_3=\dfrac{a_2}{3}=\dfrac{-1}{6}\\ a_4=\dfrac{a_3}{4}=\dfrac{-1}{24}\\ a_5=\dfrac{a_4}{5}=\dfrac{-1}{120}$ $\\$ Hence, the first five terms of the sequence are $ -1,\dfrac{-1}{2},\dfrac{-1}{6},\dfrac{-1}{24}$ and $ \dfrac{-1}{120}$ $\\$ The corresponding series is $(-1)+(\dfrac{-1}{2})+(\dfrac{-1}{6})+(\dfrac{-1}{24})+(\dfrac{-1}{120})+.......$

13   Write the first five terms of the following sequence and obtain the corresponding series: $\\$ $a_1=a_2=2,a_n=a_{n-1}-1,n > 2 $

Solution :

$a_1=a_2=2,a_n=a_{n-1}-1,n > 2 $ $\\$ $\implies a_3=a_2-1=2-1=1\\ a_4=a_3-1=1-1=0\\ a_5=a_4-1=0-1=-1$ $\\$ Hence, the first five terms of the sequence are 2, 2, 1, 0 and - 1 . $\\$ The corresponding series is 2 + 2 + 1 + 0 ( - 1 )+ ....

14   The Fibonacci sequence is defined by $1= a _1= a _2$ and $a _n = a _{n - 1 }+ a _{n - 2} , n > 2$$\\$ Find $\dfrac{a_{n+1}}a_n$,for $n=1,2,3,4,5$

Solution :

$1=a_1=a_2\\ a_n=a-{n-1}+a_{n-2},n>2\\ \therefore a_3=a_2+a_1=1+1=2\\ a_ 4 = a _3 -a _2 = 2 + 1 = 3\\ a _5 = a _4 -a _3 = 3 + 2 = 5\\ a _6 = a _5 - a_ 4 =5 + 3 = 8\\ \therefore For \ \ n=1,\dfrac{a_{n+1}}{a_n}=\dfrac{a_2}{a_1}=\dfrac{1}{1}=1\\ For \ \ n=2,\dfrac{a_{n+1}}{a_n}=\dfrac{a_3}{a_2}=\dfrac{2}{1}=2\\ For \ \ n=3,\dfrac{a_{n+1}}{a_n}=\dfrac{a_4}{a_3}=\dfrac{3}{2}\\ For \ \ n=4,\dfrac{a_{n+1}}{a_n}=\dfrac{a_5}{a_4}=\dfrac{5}{3}\\ For \ \ n=5,\dfrac{a_{n+1}}{a_n}=\dfrac{a_6}{a_5}=\dfrac{8}{5}$

15   Find the sum of odd integers from 1 to 2001.

Solution :

The odd integers from 1 to 2001 are 1, 3, 5 ...... 1999, 2001. $\\$ This sequence forms an A.P.$\\$ Here, first term, a = 1$\\$ Common difference, d = 2$\\$ Here, a +( n - 1 ) d = 2001$\\$ $= 1 +( n - 1 )( 2 )= 2001\\ = 2 _n - 2 = 2000\\ = n = 1001\\ S_n=\dfrac{n}{2}[2a+(n-1)d]\\ \therefore S_n =\dfrac{1001}{2}[2*1+(1001-1)*2]\\ =\dfrac{1001}{2}[2+1000*2]\\ =1001*1001\\ 1002001$ thus , the sum of odd numbers from 1 to 2001 is 1002001.

16   Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

Solution :

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, .... 995.$\\$ Here, a = 105 and d = 5$\\$ $a +( n - 1 ) d = 995\\ = 105 +( n - 1 ) 5 = 995\\ =( n -1 )5 = 995 -105 = 890\\ = n - 1 = 178\\ = n = 179\\ \therefore S_n=\dfrac{179}{2}[2(105)+(179-1)(5)]\\ =\dfrac{179}{2}[2(105)+(178)(5)]\\ =179[105+(89)5]\\ =179(105+445)\\ =(179)(550)\\ =98450$ $\\$ Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, 98450.

17   In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that $20 ^{th}$ term is - 112 .

Solution :

First term = 2$\\$ Let d be the common different of the A.P.$\\$ Therefore, the A.P. is 2, 2 + d , 2 + 2 d , 2 + 3 d ...$\\$ Sum of first five terms =10 +10d$\\$ Sum of next five terms = 10 + 35d$\\$ According to the given condition,$\\$ $10+10d=\dfrac{1}{4}(10+35d)\\ =40+40d=10+35d\\ =30=-5d\\ =d=-6\\ \therefore d_{20}=a+(20-1)d=2+(19)(-6)=2-114=-112$ $\\$ Thus, the 20 th of the A.P. is - 112 .

18   How many terms of the $A.P. - 6, - \dfrac{11}{2}, - 5 , ..... $are needed to give the sum -25?

Solution :

Let the sum of n terms of the given A.P. be - 25 .$\\$ It is known that,$\\$ $S_n =\dfrac{n}{2}[ 2 a +( n - 1 ) d ]$$\\$ Where n = number of terms, a = first term, and d = common difference Here, a =- 6$\\$ $d=-\dfrac{11}{2}+6=\dfrac{-11+12}{2}=\dfrac{1}{2}$$\\$ Therefore, we obtain $\\$ $-25=\dfrac{n}{2}[2*(-6)+(n-1)(\dfrac{1}{2})]\\ =-50=n[-12+\dfrac{n}{2}-\dfrac{1}{2}]\\ =-50=n[-\dfrac{25}{2}+\dfrac{n}{2}]\\ =-100=n(-25+n)\\ =n^2-25n+100=0\\ =n^2-5n-20n+100=0\\ =n(n-5)-20(n-5)=0\\ =n=20 or 5 $

19   In an A.P., if $p ^{th}$ term is 1/q and $q ^{th}$ term is 1/p, prove that the sum of first pq terms is $\dfrac{1}{2}(pq+1),$ where p$\neq $q

Solution :

It is known that the general term of an A.P. is $a_ n = a + ( n - 1 ) d $ $\\$ $\therefore $ According to the given information,$\\$ $p^{th}$ term = $a_ p = a +( p - 1 ) d =\dfrac{1}{q}.....(1)$ $\\$ $q^{th}$ term =$ a_ q = a +( q - 1 )d =\dfrac{1}{p}.....(2)$ $\\$ Subtracting (2) from (1), we obtain$\\$ $(p-1)d-(q-1)d=\dfrac{1}{q}-\dfrac{1}{p}\\ =(p-1-q+1)d=\dfrac{p-q}{pq}\\ =(p-q)d=\dfrac{p-q}{pq}\\ =d=\dfrac{1}{pq } $ $\\$ Putting the value of d in (1), we obtain$\\$ $a+(p-1)\dfrac{1}{pq}=\dfrac{1}{q}\\ =a=\dfrac{1}{q}-\dfrac{1}{q}+\dfrac{1}{pq}=\dfrac{1}{pq}\\ \therefore S_{pq}=\dfrac{pq}{2}[2a+(pq-1)d]\\ =\dfrac{pq}{2}[\dfrac{2}{pq}+(pq-1)\dfrac{1}{pq}]\\ =1+\dfrac{1}{2}(pq-1)\\ =\dfrac{1}{2}pq+1-\dfrac{1}{2}=\dfrac{1}{2}pq+\dfrac{1}{2}\\ =\dfrac{1}{2}(pq+1)$ $\\$ Thus, the sum of first pq terms of the A.P. is =$\dfrac{1}{2}(pq+1)$

20   If the sum of a certain number of terms of the A.P. 25, 22, 19, ...... is 116. Find the last term

Solution :

Let the sum of n terms of the given A.P. be 116.$\\$ $S_n=\dfrac{n}{2}[2a+(n-1)d]\\ Here,\ \ a=25 \ and \ \ d=22-25=-3\\ \therefore S_n=\dfrac{n}{2}[2*25+(n-1)(-3)]\\ =116=\dfrac{n}{2}[50-3n+3]\\ =232=n(53-3n)=53n-3n^2\\ =3n^2-53n+232=0\\ =3n^2-24n-29n+232=0\\ =3n(n-8)-29(n-8)=0\\ =(n-8)(3n-29)=0\\ =n=8 or n=\dfrac{29}{3}$$\\$ However, n cannot be equal to $\dfrac{29}{3}$ therefore, n=8$\\$ $\therefore a_8= Last \ term \ =a+(n-1)d=25+(8-1)(-3)\\ =25+(7)(-3)=25-21\\ =4$ $\\$ Thus, the last term of the A.P. is 4.

21   Find the sum to n terms of the A.P., whose $k^{ th}$ term is 5 k + 1 .

Solution :

It is given that the $k^{ th}$ term of the A.P. is 5 k + 1 .$\\$ $k^{ th}$ term =$ a_ k+( k - 1 ) d\\ \therefore a +( k - 1 ) d = 5 k + 1\\ a + kd - d = 5 k + 1$$\\$ $\therefore$ Comparing the coefficient of k, we obtain d = 5;$\\$ $= a - d = 1\\ = a - 5 = 1\\ = a = 6\\ S_n=\dfrac{n}{2}[2a+(n-1)d]\\ =\dfrac{n}{2}[2(6)+(n-1)(5)]\\ =\dfrac{n}{2}[12+5n-5]\\ =\dfrac{n}{2}[5n+7]$

22   If the sum of n terms of an A.P. is (pn + qn $^2$) , where p and q are constants, find the common difference.

Solution :

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

23   The sums of n terms of two arithmetic progressions are in the ratio 5 n + 4:9 n + 6 . Find the ratio of their 18 th terms.

Solution :

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,$\\$ $\dfrac{\text{Sumof n terms of first A.P.}}{\text{Sumof n terms of second A.P.}}=\dfrac{5n+4}{9n+6}\\ =\dfrac{\dfrac{n}{n}[2a_1+(n-1)d_1]}{\dfrac{n}{n}[2a_2+(n-1)d_2]}=\dfrac{5n+4}{9n+6}\\ =\dfrac{[2a_1+(n-1)d_1]}{[2a_2+(n-1)d_2]}=\dfrac{5n+4}{9n+6}............(1)$ $\\$ Substituting n = 35 in (1), we obtain $\\$ $\dfrac{2a_1+34d_1}{2a_2+34d_2}=\dfrac{5(35)+4}{9(35)+6}\\ =\dfrac{a_1+17d_1}{a_2+17d_2}=\dfrac{179}{321}.....(2)\\ \dfrac{18^{th}\text{termof first}}{18^{th}\text{termof second A.P.}}=\dfrac{a_1+17d_1}{a_2+17d_2}..........(3)$ $\\$ From (2) and (3), we obtain$\\$ $\dfrac{18^{th}\text{termof first}}{18^{th}\text{termof second A.P.}}=\dfrac{179}{321}$ $\\$ Thus, the ratio of 18$^{ th}$ term of both the A.P.s is 179 : 321.

24   If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first ( p + q ) terms.

Solution :

Let a and d be the first term and the common difference of the A.P. respectively.$\\$ Here,$\\$ $S_p=\dfrac{p}{2}[2a+(p-1)d]\\ S_q=\dfrac{p}{2}[2a+(q-1)d]$ $\\$ According to the given condition, $\\$ $\dfrac{p}{2}[2a+(p-1)d]=\dfrac{q}{2}[2a+(q-1)d]\\ =p[2a+(p-1)d]=q[2a+(q-1)d]\\ =2ap+pd(p-1)=2aq+qd(q-1)\\ =2a(p-q)+d[p(p-1)-q(q-1)]=0\\ =2a(p-q)+d[p^2-p-q^2+q]=0\\ =2a(p-q)+d[(p-q)(p+q)-(p-q)]=0\\ =2a(p-q)+d[(p-q)(p+q-1)]=0\\ =2a+d(p+q-1)=0\\ =d=\dfrac{-2a}{p+q-1}......(1)\\ \therefore S(p+q)=\dfrac{p+q}{2}[2a+(p+q-1).d]\\ =S_{p+q}=\dfrac{p+q}{2}[2a+(p+q-1)(\dfrac{-2a}{p+q-1})] \ \ \ [From(1)]\\ =\dfrac{p+q}{2}[2a-2a]\\ =0$ $\\$ Thus, the sum of the first ( p + q ) terms of the A.P is 0.

25   Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that$\\$ $\dfrac{a}{p}(q-r)+\dfrac{b}{q}(r-p)+\dfrac{c}{r}(p-q)=0$

Solution :

Let a 1 and d be the first term and the common difference of the A.P. respectively.$\\$ According to the given information,$\\$ $S_p=\dfrac{p}{2}[2a_1+(p-1)d]=a\\ =2a_1+(p-1)d=\dfrac{2a}{p}......(1)\\ S_q=\dfrac{q}{2}[2a_1+(q-1)d]=a\\ =2a_1+(q-1)d=\dfrac{2a}{q}......(2)\\ S_r=\dfrac{r}{2}[2a_1+(r-1)d]=a\\ =2a_1+(r-1)d=\dfrac{2a}{r}......(3)$ $\\$ Subtracting (3) from (2), we obtain$\\$ $(q-1)d-(r-1)d=\dfrac{2b}{q}-\dfrac{2c}{r}\\ =d(q-1-r+1)=\dfrac{2b}{q}-\dfrac{2c}{r}\\ =d(q-r)=\dfrac{2br-2qc}{qr}\\ =d=\dfrac{2(br-qc)}{qr(q-r)}............(5)$ $\\$ Equating both the values of d obtained in (4) and (5), we obtain $\\$ $\dfrac{aq-bp}{pq(p-q)}=\dfrac{br-qc}{qr(q-r)}\\ =qr(q-r)(aq-bq)=pq(q-q)(br-qc)\\ =r(aq-bp)(q-r)=p(br-qc)(p-q)\\ =(aqr-bpr)(q-r)=(bpr-pqc)(p-q)$ $\\$ Dividing both sides by pqr, we obtain$\\$ $(\dfrac{a}{p}-\dfrac{b}{q})(q-r)=(\dfrac{b}{q}-\dfrac{c}{r})(p-q)\\ =\dfrac{a}{p}(q-r)-\dfrac{b}{q}(q-r+p-q)+\dfrac{c}{r}(p-q)=0$ $\\$ Thus, the given result is proved.

26   The ratio of the sums of m and n terms of an A.P. is m $^2$ : n$^ 2$ . Show that the ratio of m$^{ th}$ and n $^{ th}$ term is ( 2 m - 1 ) : ( 2 n - 1 ).

Solution :

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

27   If the sum of n terms of an A.P. of $3 n^ 2 + 5 n$ and its m $^{th}$ term is 164, find the value of m.

Solution :

Let a and b be the first term and the common difference of the A.P. respectively.$\\$ $a_m=a+(m-1)d=164.......(1)$ $\\$ Sum of n terms:$S_n=\dfrac{n}{2}[2a+(n-1)d]\\ Here, \ \ \dfrac{n}{2}[2a+nd-d]=3n^2+5n\\ =na+n^2.\dfrac{d}{2}-\dfrac{nd}{2}=3n^2+5n$ $\\$ Comparing the coefficient of $n ^2$ on both sides, we obtain $\\$ $\dfrac{d}{2}=3\\ =d=6$ $\\$ Comparing the coefficient of n on both sides, we obtain $\\$ $a-\dfrac{d}{2}=5\\ =a-3=5\\ =a=8$ $\\$ 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.

28   Insert five numbers between 8 and 26 such that resulting sequence is an A.P.

Solution :

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$$\\$ $A _2 = a + 2 d = 8 + 2 * 3 = 8 + 6 = 14$$\\$ $A _3 = a + 3 d = 8 + 3 * 3 = 8 + 9 =17$$\\$ $A _4 = a + 4 d = 8 + 4 * 3 = 8 + 12 = 20$$\\$ $A _5 = a + 5 d = 8+ 5 * 3 = 8 + 15 = 23$$\\$ Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20 and 23.

29   If $\dfrac{a^n+b^n}{a^{n-1}+b^{n-1}}$ is the A.M. between a and b, then find the value of n.

Solution :

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

=$(\dfrac{a}{b})^{n-1}=1=(\dfrac{a}{b})^0\\ =n-1=0\\ =n=1$

30   Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7$^{ th}$ and ( m - 1 )$^th$ numbers is 5:9. Find the value of m.

Solution :

Let $A _1 , A _2 ,..... A _m$ be m numbers such that 1, $A _1 , A _2 ,..... A _m $,31 is an A.P.$\\$ Here, a = 1, b = 31, n = m + 2$\\$ $\therefore $ $31 = 1 +( m + 2 - 1) (d )\\ = 30 =( m + 1 ) d\\ =d=\dfrac{30}{m+1}........(1)\\ A_1=a+d\\ A_2=a+2d\\ A_3=a+3d\\ \therefore A_7=a+7d\\ A_{m-1}=a+(m-1)d$ $\\$ According to the given condition,$\\$ $\dfrac{a+7d}{a+(m-1)d}=\dfrac{5}{9}\\ =\dfrac{1+7(\dfrac{\dfrac{30}{(m+1)}})}{1+(m-1)(\dfrac{30}{m+1})} =\dfrac{5}{9} \ \ \ \ \ [From(1)]\\ =\dfrac{m+1+7(30)}{m+1+30(m-1)}=\dfrac{5}{9}\\ =\dfrac{m+1+210}{m+1+30m-30)}=\dfrac{5}{9}\\ =\dfrac{m+211}{31m-29}=\dfrac{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.

31   A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount he will pay in the 30 $^{th}$ installment?

Solution :

The first installment of the load is Rs. 100.$\\$ The second installment of the load is Rs. 105 and so on.$\\$ The amount that the man repays every month forms an A.P.$\\$ The A.P. is 100, 105, 110 ...$\\$ 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 th installment is Rs. 245.$\\$

32   The difference between any two consecutive interior angles of a polygon is 5$^o$ . If the smallest angle is 120$^o$ , find the number of the sides of the polygon.

Solution :

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

33   Find the $20^{th}$ and $n^{th}$ terms of the $ G.P.\dfrac{5}{2},\dfrac{5}{4},\dfrac{5}{8},......$

Solution :

The given G.P. is $.\dfrac{5}{2},\dfrac{5}{4},\dfrac{5}{8},......$$\\$ Here,a=First term=$\dfrac{5}{2}$ $\\$ r=Common ratio =$\dfrac{5/4}{5/2}=\dfrac{1}{2}$ $a_{20}=ar^{20-1}=\dfrac{5}{2}(\dfrac{1}{2})^{19}=\dfrac{5}{(2)(2)^{19}}=\dfrac{5}{(2)^{20}}\\ a_n=ar^{20 -1}=\dfrac{5}{2}(\dfrac{1}{2})^{19}=\dfrac{5}{(2)(2)^{19}}=\dfrac{5}{(2)^{20}}\\ a_n=ar^{n-1}=\dfrac{5}{2}(\dfrac{1}{2})^{n-1}=\dfrac{5}{(2)(2)^{n-1}}=\dfrac{5}{(2)^n}$

34   Find the 12 $^{th}$ term of a G.P. whose 8 $^{th}$ term is 192 and the common ratio is 2.

Solution :

Common ratio, r = 2$\\$ Let a be the first term of the G.P.$\\$ $\therefore a_8=ar^{8-1}=ar^7=ar^7=192=a(2)^7=192=a(7)^7=(2)^6(3)\\ =a=\dfrac{(2)^6*3}{(2)^7}=\dfrac{3}{2}\\ \therefore a_{12}=ar^{12-1}=(\dfrac{3}{2})(2)^{11}=(3)(2)^{10}=3072$

35   The 5$^{ th}$ , 8 $^{ th}$ and 11 $^{ th}$ terms of a G.P. are p , q and s , respectively. Show that q$^2$ = ps .

Solution :

Let a be the first term and r be the common ratio of the G.P. According to the given condition,$\\$ $a_5=ar^{5-1}=ar^4=p..............(1)\\ a_8=ar^{8-1}=ar^7=q............(2)\\ a_{11}=ar^{11-1}=ar^{10}=s........(3)$ $\\$ Dividing equation (2) by (1), we obtain $\\$ $\dfrac{ar^7}{ar^4}=\dfrac{q}{p}\\ r^3=\dfrac{q}{p}.......(4)$ $\\$ Dividing equation (3) by (2), we obtain$\\$ $\dfrac{ar^{10}}{ar^7}=\dfrac{s}{q}\\ =r^3=\dfrac{s}{q}.......(5)$ $\\$ Equating the values of r $^3$ obtained in (4) and (5), we obtain$\\$ $\dfrac{q}{p}=\dfrac{s}{q}\\ =q^2=ps$ $\\$ Thus, the given result is proved.

36   The 4 $^{th}$ term of a G.P. is square of its second term, and the first term is - 3 . Determine its 7 $^{th}$ term.

Solution :

Let a be the first term and r be the common ratio of the G.P. $\\$ $\therefore a =- 3$ $\\$ It is known that, $a _n = ar ^{n - 1}$ $\\$ $\therefore a _4 =ar^ 3 =(- 3 ) r ^3\\ a_ 2 = a r^ 1 =(- 3 ) r$ $\\$ According to the given condition,$\\$ $(- 3 ) r^ 3 =[(-3 ) r]^2\\ =- 3 r^ 3=9 r^ 2 = r =-3 a_ 7 = a r ^{7 - 1 } =ar^ 6 =(- 3 )(- 3 )6 =-(3)^7=- 2187$ Thus, the seventh term of the G.P. is - 2187 .

37   The 4 $^{th}$ term of a G.P. is square of its second term, and the first term is - 3 . Determine its 7 $^{th}$ term.

Solution :

Let a be the first term and r be the common ratio of the G.P. $\therefore a =- 3$ $\\$ It is known that, $a_n = ar^{ n - 1}\\ \therefore a_4 = ar ^3 =(- 3) r^ 3\\ a_2 = a r^ 1 =(- 3 ) r$ $\\$ According to the given condition, $\\$ $(- 3 ) r ^3 =[(- 3 ) r ]^ 2\\ =- 3 r ^3 = 9 r^ 2 = r =- 3 a_ 7 = a r ^{7 - 1 }= ar ^6 =(- 3 )(- 3 )^6=-( 3 )^7=- 2187$ $\\$ Thus, the seventh term of the G.P. is - 2187 .

38   Which term of the following sequences:$\\$ (a) 2, 2$\sqrt{ 2}$, 4..... is 128?$\\$ (b) 3,3,3 3,...... is 729?$\\$ (c)$\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{27},..... is \dfrac{1}{19683}?$

Solution :

(a) The given sequence is 2, 2 $\sqrt{2}$, 4..... is 128? $\\$ Here, a = 2 and r = (2$\sqrt{ 2 })/ 2 =\sqrt{2}$ Let the n$^{ th}$ term of the given sequence be 128.$\\$ $a_ n = a r^{ n - 1}\\ =(2)(\sqrt{2})^{n-1}=128\\ =(2)(2)^{\frac{n-1}{2}}=(2)^7\\ =(2)^{\frac{n-1}{2}+1}=(2)^7\\ \therefore \dfrac{n-1}{2}+1=7\\ =\dfrac{n-1}{2}=6\\ =n-1=12\\ =n=13$ $\\$ Thus, the 13$^{ th}$ term of the given sequence is 128.

(b) The given sequence is $\sqrt{3},3,3\sqrt{ 3},......$ $\\$ $a=\sqrt{3}$ and $ r=\dfrac{3}{\sqrt{3}}=\sqrt{3}$ $\\$ Let the n$^{ th}$ term of the given sequence be 729. $\\$ $a _n = a r^{ n -1}\\ \therefore a r^{ n - 1 }= 729\\ =(\sqrt{3})(\sqrt{3})^(n-1)=729\\ =(3)^{\frac{1}{2}}(3)^{\frac{n-1}{2}}=(3)^6\\ =(3)^{\frac{1}{2}+\frac{n-1}{2}}=(3)^6\\ \therefore \dfrac{1}{2}+\dfrac{n-1}{2}=6\\ =\dfrac{1+n-1}{2}=6\\ =n=12$ $\\$ Thus, the 12$^{ th}$ term of the given sequence is 729.

39   For what values of x, the numbers $\dfrac{2}{7},x,-\dfrac{7}{2}$ are in G.P.?

Solution :

The given numbers are $ \dfrac{-2}{7},x,\dfrac{-7}{2}$ $\\$ Common ratio=$ \dfrac{x}{-2/7}=\dfrac{-7x}{2}$ $\\$ Also, common ratio=$ \dfrac{-7/2}{x}=\dfrac{-7}{2x}$ $\\$ $\therefore \dfrac{-7x}{2}=\dfrac{-7}{2x}\\ =x^2=\dfrac{-2*7}{-2*7}=1\\ =x=\sqrt{1}\\ =x=\pm 1$ $\\$ Thus, for $x =\pm 1$ , the given numbers will be in G.P.

40   Find the sum up to 20 terms in the geometric progression 0.15, 0.015, 0.0015....

Solution :

The given G.P. is 0.15, 0.015, 0.00015 ...$\\$ Here,a=0.15 and r=$\dfrac{0.015}{0.15}=0.1$ $\\$ $S_n=\dfrac{a(1-r^n)}{1-r}\\ \therefore S_{20}=\dfrac{0.15[1-(0.1)^{20}]}{1-0.1}\\ =\dfrac{0.15}{0.9}[1-(0.1)^{20}]\\ =\dfrac{15}{90}[1-(0.1)^{20}]\\ =\dfrac{1}{6}[1-(0.1)^{20}]$ $\\$

41   Find the sum of n terms in the geometric progression $\sqrt{7},\sqrt{21},3\sqrt{7},......$

Solution :

The given G.P. is $\sqrt{7},\sqrt{21},3\sqrt{7},.....$ $\\$ Here,$ a=\sqrt{7}$ and $r=\dfrac{\sqrt{21}}{7}=\sqrt{3}\\ S_n=\dfrac{a(1-r^n)}{1-r}\\ =S_n=\dfrac{\sqrt{7}[1-(\sqrt{3})^n]}{1-\sqrt{3}}\\ =S_n=\dfrac{\sqrt{7}[1-(\sqrt{3})^n]}{1-\sqrt{3}}*\dfrac{1+\sqrt{3}}{1+\sqrt{3}}\\ =S_n=\dfrac{\sqrt{7}(\sqrt{3}+1)[1-(\sqrt{3})^n]}{1-3}\\ =S_n=\dfrac{-\sqrt{7}(\sqrt{3+1})[1-(\sqrt{3})^n]}{2}\\ =\dfrac{\sqrt{7}(1+\sqrt{3})}{2}[(3)^{\frac{n}{2}}-1]$ $\\$

42   Find the sum of n terms in the geometric progression 1, -a , a $^2$ , - a $^3$ ..... ( if a $\neq - 1 )$

Solution :

The given G.P. is $1, - a , a ^2 , - a^ 3 .....$$\\$ Here, first term =$ a _1 = 1$ $\\$ Common ratio = r =- a$\\$ $S_n=\dfrac{a_1(1-r^n)}{1-r}\\ \therefore S_n=\dfrac{1[1-(-a)^n]}{1-(-a)}=\dfrac{[1-(-a)^n]}{1+a}$

43   Evaluate $\displaystyle\sum_{k=1}^{11} (2+3^k)$

Solution :

$\displaystyle\sum_{k=1}^{11} (2+3^k)=\displaystyle\sum_{k=1}^{11}(2)+\displaystyle\sum_{k=1}^{11}(3^k)=22+\displaystyle\sum_{k=1}^{11}3^k .........(1)\\ \displaystyle\sum_{k=1}^{11}3^k=3^1+3^2+3^3+.....+3^{11}$ $\\$ The terms of this sequence$3,3 ^2 ,3^ 3 ...... $ forms a G.P.$\\$ $S_n=\dfrac{a(r^n-1)}{r-1}\\ S_n=\dfrac{3[(3)^{11}-1]}{[3-1]}\\ S_n=\dfrac{3}{2}(3^{11}-1)\\ \therefore \displaystyle\sum_{k=1}^{11} 3^k=\dfrac{3}{2}(3^{11}-1)$ $\\$ Substituting this value in equation (1), we obtain$\\$ $\displaystyle\sum_{k=1}^{11}(2+3^k)=22+\dfrac{3}{2}(3^{11}-1)$

44   The sum of first three terms of a G.P. is $\dfrac{39}{10}$ and their product is 1. Find the common ratio and the terms.

Solution :

Let $\dfrac{a}{r},a,ar$ be the first three terms of the G.P.$\\$ $\dfrac{a}{r}+a+ar=\dfrac{39}{10} ......(1)\\ (\dfrac{a}{r})(a)(ar)=1......(2)$ $\\$ From (2), we $\\$ Obtain $a^ 3 = 1$ $\\$ $=a=1$(Considering real roots only) $\\$ Substituting a = 1 in equation (1), we obtain $\\$ $\dfrac{1}{r}+1+r=\dfrac{39}{10}\\ =1+r+r^2=\dfrac{39}{10}r\\ =10+10r+10r^2-39r=0\\ =10r^2-29r+10=0\\ =10r^2-25r-4r+10=0\\ =5r(2r-5)-2(2r-5)=0\\ =(5r-2)(2r-5)=0\\ =r=\dfrac{2}{5} or \dfrac{5}{2}$ $\\$ Thus, the three terms of G.P. are $\dfrac{5}{2},1 $ and $ \dfrac{2}{5}.$

45   How many terms of $G.P. 3,3\sqrt{ 2} ,3 \sqrt{3} ...$ are needed to give the sum 120?

Solution :

The given G.P. is $3,3\sqrt{ 2} ,3\sqrt{ 3 }...$ $\\$ Let n terms of this G.P. be required to obtain in the sum as 120.$\\$ $ S_n=\dfrac{a(1-r^n)}{1-r}$ $\\$ Here,$a=3 $ and $r=3$ $\\$ $\therefore S_n=120=\dfrac{(3^n-1)}{3-1}\\ =120=\dfrac{3(3^n-1)}{2}\\ =\dfrac{120*2}{3}=3^n-1\\ =3^n-1=80\\ =3^n=81\\ =3^n=3^4\\ \therefore n=4$ $\\$ Thus, four terms of the given G.P. are required to obtain the sum as 120.

46   The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Solution :

Let the G.P. be $a , ar , ar^ 2 , ar^ 3 ,....$ According to the given condition,$\\$ $a + ar + ar^ 2 = 16$ and $ar^ 3 + ar^ 4 + ar^ 5 = 128\\ =a(1+r+r^3)=16........(1)\\ ar^3(1+r+r^2)=128.....(2)$ $\\$ Dividing equation (2) by (1), we obtain$\\$ $\dfrac{ar^3(1+r+r^3)}{a(1+r+r^2)}=\dfrac{128}{16}\\ =r^3=8\\ \therefore r=2$ $\\$ Substituting r = 2 in (1), we obtain a ( 1 + 2 + 4 )= 16$\\$ $=a(7)=16\\ =a=\dfrac{16}{7}\\ S_n=\dfrac{a(r^n-1)}{r-1}\\ =S_n=\dfrac{16}{7}\dfrac{(2^n-1)}{2-1}=\dfrac{16}{7}(2^n-1)$

47   Given a G.P. with a = 729 and 7$^{ th}$ term 64, determine $S_ 7$ .

Solution :

$a = 729 \ \ a _7 = 64$ $\\$ Let r be the common ratio of the G.P. It is known that, $\\$ $a _n = a r^{ n - 1}\\ a_7 = ar^{ 7 - 1} =( 729 ) r^ 6\\ = 64 = 729r^ 6\\ =r^6=(\dfrac{2}{3})^6\\ =r=\dfrac{2}{3}$ $\\$ Also, it is known that, $\\$ $ S_n=\dfrac{a(1-r^n)}{1-r}\\ \therefore S_7=\dfrac{729(1-(\dfrac{2}{3})^7)}{1-\dfrac{2}{3}}\\ =3*729[1-(\dfrac{2}{3})^7]\\ =(3)^7[\dfrac{(3)^7-(2)^7}{(3)^7}]\\ =(3)^7-(2)^7\\ =2187-128\\ =2059$

48   Find a G.P. for which sum of the first two terms is - 4 and the fifth term is 4 times the third term.

Solution :

Let a be the first term and r be the common ratio of the G.P.$\\$ According to the given conditions,$\\$ $A_2=-4=\dfrac{a(1-r^2)}{1-r}......(1)\\ a_5=4*a_3\\ =ar^4=4ar^2=r^2=4\\ \therefore r=\pm 2$ $\\$ From (1), we obtain $\\$ $-4=\dfrac{a[1-(2)^2]}{1-2}$ for r=2 $\\$ $=-4=\dfrac{a(1-4)}{-1}\\ =-4=a(3)\\ =a=\dfrac{-4}{3}$ $\\$ Also,$-4=\dfrac{a[1-(-2)^2]}{1-(-2)}$ for r=-2 $\\$ $=-4 =\dfrac{a(1-4)}{1+2}\\ =-4=\dfrac{a(-3)}{3}\\ =a=4$ $\\$ Thus, the required G.P. is $\dfrac{-4}{3},\dfrac{-8}{3},\dfrac{-16}{3},.... or 4,-8,-16,-32.....$

49   If the 4$^{ th}$ , 10 $^{ th}$ and 16 $^{ th}$ terms of a G.P. are x , y , and z, respectively. Prove that x , y , z are in G.P.

Solution :

Let a be the first term and r be the common ratio of the G.P.$\\$ According to the given condition,$\\$ $a _4 = a r^ 3 = x ...... ( 1 )\\ a^{ 10 }= a r ^9 = y ...... ( 2 )\\ a ^{16 }= a r^{ 15}= z ...... ( 3 )$ $\\$ Dividing (2) by (1), we obtain$\\$ $\dfrac{y}{x}=\dfrac{ar^9}{ar^3}=\dfrac{y}{x}=r^6$ $\\$ Dividing (3) by (2), we obtain $\\$ $\dfrac{z}{y}=\dfrac{ar^{15}}{ar^9}=\dfrac{z}{y}=r^6\\ \therefore \dfrac{y}{x}=\dfrac{z}{y}$ $\\$ Thus, x , y , z are in G.P.

50   Find the sum to n terms of the sequence, 8, 88, 888, 8888 ....

Solution :

The given sequence is 8, 88, 888, 8888 ....$\\$ This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as$\\$ $S_n=8+88+888+8888+.....$ to n terms$\\$ $\dfrac{8}{9}[9+99+999+9999+......$ to n terms]$\\$ $\dfrac{8}{9}[(10-1)+(10^2-)+(10^3-1)+(10^4-1)+..........$ to n terms]$\\$ $\dfrac{8}{9}[(10+10^2+......$n terms)-($1+1+1+.....$ n term)]$\\$ $\dfrac{8}{9}[\dfrac{10(10^n-1)}{10-1}-n]\\ =\dfrac{8}{9}[\dfrac{10(10^n-1)}{9}-n]\\ =\dfrac{80}{81}(10^n-1)-\dfrac{8}{9}n$ $\\$

51   Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.

Solution :

Required sum = $2 * 128 + 4 * 32 + 8 * 8 + 16 * 2 + 32 *\dfrac{1}{2}$ $\\$ $=64[4+2+1+\dfrac{1}{2}+\dfrac{1}{2^2}]$ $\\$ Here, 4, 2,1,$\dfrac{1}{2},\dfrac{1}{2^2}$ is a G.P.$\\$ First term, a = 4$\\$ Common ratio,$r=\dfrac{1}{2}$ $\\$ It is known that, $ S_n=\dfrac{a(1-r^n)}{1-r}\\ \therefore S_5=\dfrac{4[1-(\dfrac{1}{2})^5]}{1-\dfrac{1}{2}}\\ =\dfrac{4[1-(\dfrac{1}{32})]}{\dfrac{1}{2}}=8(\dfrac{32-1}{32})=\dfrac{31}{4}$ $\\$ $\therefore $ Required sum $ =64(\dfrac{31}{4})=(16)(31)=496$

52   Show that the products of the corresponding terms of the sequences form a , ar , ar $^2$ ,..... ar$^ {n - 1}$ and $A , AR , AR ^2 , AR ^{n - 1 }$a G.P., and find the common ratio.

Solution :

It has to be proved that the sequence: $aA , arAR , ar ^2 AR ^2 , ....... ar^{ n - 1} AR ^{n - 1}$ , forms a G.P. $\\$ $\dfrac{\text{Second term}}{\text{First term}}=\dfrac{ar \ AR}{aA}=rR\\ \dfrac{\text{Third term}}{\text{Second term}}=\dfrac{ar^2 \ AR^2}{aA}=rR$ Thus, the above sequence forms a G.P. and the common ratio is rR.

53   Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4$^{ th}$ by 18.

Solution :

Let a be the first term and r be the common ratio of the G.P.$\\$ $a _1 = a , a _2 = ar , a _3 =ar^ 2 , a_ 4 = ar ^3$$\\$ By the given condition,$\\$ $a _3 = a _1 + 9 = ar^ 2 = a + 9 ..... ( 1 )\\ a_ 4 = a _4 + 18 = ar = ar^ 3 + 18 ..... ( 2 )$$\\$ From (1) and (2), we obtain$\\$ $a(r^2-1)=9...........(3)\\ ar(1-r^2)=18..........(4)$ $\\$ Dividing (4) by (3),we obtain $\\$ $\dfrac{ar(1-r^2)}{a(r^2-1)}=\dfrac{18}{9}\\ =-r=2\\ r=-2$ $\\$ Substituting the value of r in(1),we obtain $\\$ $4a=a+9\\ =3a=9\\ \therefore a=3 $ $\\$ Thus, the first four numbers of the G.P. are $3, 3 (- 2 ) , 3 (- 2 )^2$ , and $3 (- 2 )^3\\ i.e., 3,-6,12,\ \ and \ \ -24.$

54   If p$^{ th}$ , q $^{ th}$ and r $^{ th}$ terms of a G.P. are a , b and c , respectively. Prove that a $^{q - r}$ . b$^{ r-p}$ . c $^{p-q}$ = 1 .

Solution :

Let A be the first term and R be the common ratio of the G.P.$\\$ According to the given information,$\\$ $AR^{ p - 1 }= a\\ AR^{ q - 1 }= b\\ AR^{ r - 1 }= c\\ a^{q-r}.b^{r-p}.c^{p-q}\\ =A^{q-r}*R^{(p-1)(q-r)}*A^{r-p}*R^{(q-1)(r-p)}*A^{p-q}*R{(r-1)(p-q)}\\ =A^{q-r+r-p+p-q}*R^{(pr-pr-q+r)+(rp-r+p-pq)+(pr-p-qr+q)}\\ =A^o*R^o\\ =1$ $\\$ Thus, the given result is proved.

55   If the first and the n $^{th}$ term of a G.P. are a and b , respectively, and if P is the product of n terms, prove that $P ^2 =( ab )^n$ .

Solution :

The first term of the G.P is a and the last term is b .$\\$ Therefore, the G.P. is $a , ar , ar ^2 , ar ^3 .... ar^{ n - 1}$ , where r is the common ratio.$\\$ $b=ar^{n-1}......(1)$$\\$ P=Product of n terms $\\$ $=(a)(ar)(ar^2).....(ar^{n-1})\\ =(a*a*....a)(r*r^2*.....r^{n-1})\\ =a^nr^{1+2+.....(n-1)}..........(2)$ $\\$ Here, 1,2.....(n-1) is an A.P.$\\$ $\therefore 1+2+....+(n-1)\\ =\dfrac{n-1}{2}[2+(n-1-1)*1]=\dfrac{n-1}{2}[2+n-2]=\dfrac{n(n-1)}{2}\\ P=a^n r^{\frac{n(n-1)}{2}}\\ \therefore P^2=a^2n r^{n(n-1)}\\ =[a^2 r^(n-1)]^n\\ =[a*ar^{n-1}]^n\\ =(ab)^n \ \ \ \ [Using (1)]$ $\\$ Thus, the given result is proved.

56   Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from ( n + 1 )$^{th}$ to $(2n)^{th}$ term is $\dfrac{1}{r^n}.$

Solution :

Let a be the first term and r be the common ratio of the G.P.$\\$ Sum of first n terms =$\dfrac{a(1-r^n)}{(1-r)}$ $\\$ Since there are n terms from $(n+1)^{th}$ to $(2n)^{th}$ term,$\\$ Sum of terms from $(n+1)^{th}$ to $(2n)^{th}$ term $\\$ $S_n=\dfrac{a_{n+1}(1-r^n)}{1-r}\\ a^{n+1}=ar^{n+1-1}=ar^n $ $\\$ Thus, required ratio =$\dfrac{a(1-r^n)}{(1-r)}*\dfrac{(1-r)}{ar^n(1-r^n)}=\dfrac{1}{r^n}$ $\\$ Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from $(n+1)^{th}$ to $(2n)^{th}$ term is $ \dfrac{1}{r^n}$

57   If a , b , c and d are in G.P. show that:$\\$ $( a^2+b ^2 + c ^2 )( b^ 2 + c^ 2 + d^ 2 )=( ab + bc + cd )^2$

Solution :

If a , b , c and d are in G.P. Therefore,$\\$ $bc = ad ........ ( 1 )\\ b^ 2 = ac ........ ( 2 )\\ c ^2 = bd ......... ( 3 )$$\\$ It has to be proved that,$\\$ $( a^2+b ^2 + c ^2 )( b^ 2 + c^ 2 + d^ 2 )=( ab + bc + cd )^2$$\\$ R.H.S.$\\$ $=(ab+bc+cd)^2\\ =(ab+ad+cd)^2\ \ \ [Using (1)]\\ [ab+d(a+c)]^2\\ =a^2b^2+2abd(a+c)+d^2(a+c)^2\\ =a^2b^2+2a^2bd+2acbd+d^2(a^2+2ac+c^2)\\ =a^2b^2+2a^2c^2+2b^2c^2+d^2a^2+2d^2b^2+d^2c^2 \ \ \ \ \ \ [Using(1)and(2)]\\ =a^2b^2+a^2c^2+a^2c^2+b^2c^2+b^2c^2+d^2a^2+d^2b^2+d^2b^2+d^2c^2\\ =a^2b^2+a^2c^2+a^2d^2+b^2*b^2+b^2c^2+b^2d^2+c^2b^2+c^2*c^2+c^2d^2$ $\\$ [Using (2) and (3) and rearranging terms]$\\$ $=a^2(b^2+c^2+d^2 )+b^2(b^2+c^2+d^2)+c^2(b^2+c^2+d^2)\\ =(a^2+b^2+c^2 )(b^2+c^2+d^2)=L.H.S\\ \therefore L.H.S=R.H.S.\\ \therefore (a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2.$

58   Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Solution :

Let $G _1$ and $G_ 2$ be two numbers between 3 and 81 such that the series, 3, $G _1$ , $G_ 2$ ,81 , forms a G.P.$\\$ Let a be the first term and r be the common ratio of the G.P.$\\$ $\therefore 81=(3)(r)^3\\ =r^3=27\\ \therefore r=3$ (Talking real roots only)$\\$ For r = 3 ,$\\$ $G _1 = ar =( 3 )( 3)= 9\\ G _2 = ar^ 2 =( 3)( 3 )^2= 27$ $\\$ Thus, the required two numbers are 9 and 27.

59   Find the value of n so that $\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}$ may be the geometric mean between a and b .

Solution :

M. of a and b is $\sqrt{ab}$ $\\$ By the given condition: $\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{ab}$$\\$ Squaring both sides, we obtain$\\$ $ \dfrac{(a^{n+1}+b^{n+1})^2}{(a^n+b^n)^2}=ab \\ =a^{2n+2}+2a^{n+1}b^{n+1}+b^{2n+2}=(ab)(a^{2n}+2a^n b^n+b^{2n})\\ =a^{2n+2}+2a^{n+1}b^{n+1}+b^{2n+2}=a^{2n+1}b+2a^{n+1}b^{n+1}+ab^{2n+1}\\ =a^{2n+2}+b^{2n+2}=a^{(2n+1)}b+ab^{2n+1}\\ =a^{2n+2}-a^{2n+1}b =ab^{2n+1}-b^{2n+2}\\ =a^{2n+1}(a-b)=b^{2n+1}(a-b)\\ =\dfrac{a}{b}^{2n+1}=1=\dfrac{a}{b}^o\\ 2n+1=0\\ 2n+1=0\\ n=\dfrac{-1}{2}$

60   The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio $( 3 + 2 \sqrt{2} ) : ( 3 - 2\sqrt{ 2} )$

Solution :

Le the two numbers be a and b .$\\$ G.M. =$\sqrt{ ab}$ $\\$ According to the given condition,$\\$ $a+b=6\sqrt{ab} ..........(1)\\ =(a+b)^2=36(ab)$ $\\$ Also,$\\$ $(a-b)^2=(a+b)^2-4ab=36ab-4ab=32ab\\ =a-b=\sqrt{32}\sqrt{ab}\\ =4\sqrt{2}\sqrt{ab}......(2)$ $\\$ Adding (1) and (2), we obtain$\\$ $2a=(6+4\sqrt{2})\sqrt{ab}\\ a=(3+2\sqrt{2})\sqrt{ab}$ $\\$ Substituting the value of a in (1), we obtain$\\$ $b=6\sqrt{ab}-(3+2\sqrt{2})\sqrt{ab}\\ =b=(3-2\sqrt{2})\sqrt{ab}\\ \dfrac{a}{b}=\dfrac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}}=\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}$ $\\$ Thus , the required ratio is $ (3+2\sqrt{2}):(3-2\sqrt{2}).$