# 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 $a_n=\dfrac{2n-3}{6}$

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 $a_n=n\dfrac{n^2+5}{4}$

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 .

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.

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.

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 $\neq - 1 )$

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 sequence3,3 ^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 $(3+2\sqrt{2}):(3-2\sqrt{2}).$