NCERT Solutions: Class 11 Maths Chapter 8 - Sequences and Series

Put $n=1,2,3,4,5$ in $a_n=n(n+2)$. This gives $1\cdot3=3$, $2\cdot4=8$, $3\cdot5=15$, $4\cdot6=24$ and $5\cdot7=35$.

Substitute $n=1,2,3,4,5$ in $a_n=\dfrac{n}{n+1}$.

For $n=1,2,3,4,5$, $2^n$ gives $2^1,2^2,2^3,2^4,2^5$, i.e. $2,4,8,16,32$.

Substitute $n=1,2,3,4,5$: $\dfrac{-1}{6},\dfrac{1}{6},\dfrac{3}{6},\dfrac{5}{6},\dfrac{7}{6}$.

For $n=1,2,3,4,5$, the sign alternates by $(-1)^{n-1}$ and the powers are $5^2,5^3,5^4,5^5,5^6$.

Put $n=1,2,3,4,5$ in $a_n=\dfrac{n(n^2+5)}{4}$. The values are $\dfrac{6}{4}$, $\dfrac{18}{4}$, $\dfrac{42}{4}$, $\dfrac{84}{4}$ and $\dfrac{150}{4}$.

$a_{17}=4(17)-3=65$ and $a_{24}=4(24)-3=93$.

$a_7=\dfrac{7^2}{2^7}=\dfrac{49}{128}$.

$a_9=(-1)^8\cdot9^3=729$.

$a_{20}=\dfrac{20(20-2)}{20+3}=\dfrac{360}{23}$.

$a_1=3$. Then $a_2=3(3)+2=11$, $a_3=3(11)+2=35$, $a_4=3(35)+2=107$ and $a_5=3(107)+2=323$.

$a_2=\dfrac{-1}{2}$, $a_3=\dfrac{-1/2}{3}=-\dfrac16$, $a_4=\dfrac{-1/6}{4}=-\dfrac1{24}$ and $a_5=\dfrac{-1/24}{5}=-\dfrac1{120}$.

$a_1=2$ and $a_2=2$. For $n>2$, subtract 1 from the preceding term: $a_3=1$, $a_4=0$, $a_5=-1$.

The Fibonacci terms are $a_1=1$, $a_2=1$, $a_3=2$, $a_4=3$, $a_5=5$, $a_6=8$. Hence $\dfrac{a_2}{a_1}=1$, $\dfrac{a_3}{a_2}=2$, $\dfrac{a_4}{a_3}=\dfrac32$, $\dfrac{a_5}{a_4}=\dfrac53$ and $\dfrac{a_6}{a_5}=\dfrac85$.

Here $a=\dfrac52$ and $r=\dfrac12$. Thus $a_n=ar^{n-1}=\dfrac52\left(\dfrac12\right)^{n-1}=\dfrac{5}{2^n}$. Hence $a_{20}=\dfrac{5}{2^{20}}$.

In a G.P., $a_{12}=a_8r^4$. Therefore $a_{12}=192\times2^4=192\times16=3072$.

Let the first term be $a$ and common ratio be $r$. Then $p=ar^4$, $q=ar^7$ and $s=ar^{10}$. Hence $q^2=a^2r^{14}$ and $ps=(ar^4)(ar^{10})=a^2r^{14}$. Therefore $q^2=ps$.

Let the common ratio be $r$. Since $a=-3$, the second term is $-3r$ and the fourth term is $-3r^3$. Given $-3r^3=(-3r)^2=9r^2$. Since terms of a G.P. are non-zero, $r\ne0$, so $r=-3$. The seventh term is $ar^6=-3(-3)^6=-2187$.

(a) Here $a=2$, $r=\sqrt2$. So $a_n=2(\sqrt2)^{n-1}=2^7$ gives $1+\dfrac{n-1}{2}=7$, hence $n=13$. (b) Here $a=\sqrt3$ and $r=\sqrt3$, so $a_n=(\sqrt3)^n=3^{n/2}$. Since $729=3^6$, $n=12$. (c) The sequence is $\left(\dfrac13\right)^n$, so $\dfrac1{19683}=\dfrac1{3^9}$ is the 9th term.

For three numbers in G.P., the square of the middle term equals the product of the other two. Thus $x^2=\left(-\dfrac27\right)\left(-\dfrac72\right)=1$, so $x=\pm1$.

Here $a=0.15=\dfrac3{20}$ and $r=0.1=\dfrac1{10}$. Therefore $S_{20}=\dfrac{a(1-r^{20})}{1-r}=\dfrac{3/20\,(1-10^{-20})}{9/10}=\dfrac16(1-10^{-20})$.

Here $a=\sqrt7$ and $r=\dfrac{\sqrt{21}}{\sqrt7}=\sqrt3$. Thus $S_n=\dfrac{a(r^n-1)}{r-1}=\dfrac{\sqrt7\left((\sqrt3)^n-1\right)}{\sqrt3-1}$.

The first term is $1$ and the common ratio is $-a$. Since $a\ne-1$, $1-(-a)=1+a\ne0$. Therefore $S_n=\dfrac{1-(-a)^n}{1+a}$.

Here $a=x^3$ and $r=x^2$. Since $x\ne\pm1$, $r\ne1$. Thus $S_n=\dfrac{x^3((x^2)^n-1)}{x^2-1}=\dfrac{x^3(x^{2n}-1)}{x^2-1}$.

$\sum_{k=1}^{11}(2+3^k)=22+\sum_{k=1}^{11}3^k=22+\dfrac{3(3^{11}-1)}{3-1}=22+265719=265741$.

Let the three terms be $\dfrac{a}{r},a,ar$. Their product is $a^3=1$, so $a=1$. The sum gives $r+1+\dfrac1r=\dfrac{39}{10}$, so $r+\dfrac1r=\dfrac{29}{10}$. Hence $10r^2-29r+10=0$, giving $r=\dfrac52$ or $r=\dfrac25$.

Here $a=3$ and $r=3$. The sum of $n$ terms is $S_n=\dfrac{3(3^n-1)}{3-1}$. Set this equal to 120: $\dfrac{3(3^n-1)}2=120$, so $3^n=81=3^4$. Hence $n=4$.

Let the sum of the first three terms be $a(1+r+r^2)=16$. The next three terms have sum $ar^3(1+r+r^2)=128$. Dividing gives $r^3=8$, so $r=2$. Then $a(1+2+4)=16$, hence $a=\dfrac{16}{7}$. Therefore $S_n=\dfrac{a(r^n-1)}{r-1}=\dfrac{16}{7}(2^n-1)$.

The 7th term is $ar^6$, so $729r^6=64$. Hence $r=\pm\dfrac23$. For $r=\dfrac23$, $S_7=\dfrac{729(1-(2/3)^7)}{1-2/3}=2059$. For $r=-\dfrac23$, $S_7=\dfrac{729(1-(-2/3)^7)}{1+2/3}=463$.

Let the first term be $a$ and common ratio be $r$. Then $a(1+r)=-4$. Also, the fifth term is 4 times the third term, so $ar^4=4ar^2$. Since $a\ne0$ and $r\ne0$, $r^2=4$, giving $r=2$ or $r=-2$. If $r=2$, then $3a=-4$, so $a=-\dfrac43$. If $r=-2$, then $-a=-4$, so $a=4$.

Let the first term be $a$ and common ratio be $r$. Then $x=ar^3$, $y=ar^9$ and $z=ar^{15}$. Now $y^2=a^2r^{18}$ and $xz=(ar^3)(ar^{15})=a^2r^{18}$. Hence $y^2=xz$, so $x,y,z$ are in G.P.

The $k^{\text{th}}$ term is $8(1+10+\cdots+10^{k-1})=\dfrac89(10^k-1)$. Therefore $S_n=\dfrac89\left(\sum_{k=1}^n10^k-n\right)=\dfrac89\left(\dfrac{10(10^n-1)}9-n\right)=\dfrac{8(10^{n+1}-10-9n)}{81}$.

The products of corresponding terms are $2\cdot128=256$, $4\cdot32=128$, $8\cdot8=64$, $16\cdot2=32$ and $32\cdot\dfrac12=16$. Their sum is $256+128+64+32+16=496$.

The corresponding products are $aA$, $(ar)(AR)=aA(rR)$, $(ar^2)(AR^2)=aA(rR)^2,\ldots,aA(rR)^{n-1}$. Each term is obtained from the previous one by multiplying by $rR$. Hence they form a G.P. with common ratio $rR$.

Let the four terms be $a,ar,ar^2,ar^3$. Then $ar^2-a=9$ and $ar-ar^3=18$. Thus $a(r^2-1)=9$ and $ar(1-r^2)=18$. Dividing the second equation by the first gives $-r=2$, so $r=-2$. Then $a(4-1)=9$, giving $a=3$. The numbers are $3,-6,12,-24$.

Let the first term be $A$ and common ratio be $R$. Then $a=AR^{p-1}$, $b=AR^{q-1}$ and $c=AR^{r-1}$. Therefore $a^{q-r}b^{r-p}c^{p-q}=A^{(q-r)+(r-p)+(p-q)}R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}$. The exponent of $A$ is $0$, and the exponent of $R$ also simplifies to $0$. Hence the product is $1$.

Let the common ratio be $r$. Since the first term is $a$ and the $n^{\text{th}}$ term is $b$, $b=ar^{n-1}$. The product of $n$ terms is $P=a^nr^{0+1+\cdots+(n-1)}=a^nr^{n(n-1)/2}$. Thus $P^2=a^{2n}r^{n(n-1)}$. Also $(ab)^n=(a\cdot ar^{n-1})^n=a^{2n}r^{n(n-1)}$. Hence $P^2=(ab)^n$.

Let the first term be $a$ and common ratio be $r$. The sum of the first $n$ terms is $S_1=\dfrac{a(r^n-1)}{r-1}$ for $r\ne1$. The sum from the $(n+1)^{\text{th}}$ to the $(2n)^{\text{th}}$ term is $S_2=ar^n\dfrac{r^n-1}{r-1}=r^nS_1$. Therefore $S_1:S_2=1:r^n$, i.e. the ratio is $\dfrac1{r^n}$.

Let $a,b,c,d$ be $a,ar,ar^2,ar^3$. Then $a^2+b^2+c^2=a^2(1+r^2+r^4)$ and $b^2+c^2+d^2=a^2r^2(1+r^2+r^4)$. Hence the left side is $a^4r^2(1+r^2+r^4)^2$. Also $ab+bc+cd=a^2r+a^2r^3+a^2r^5=a^2r(1+r^2+r^4)$, whose square is the same expression.

Let the four terms be $3, 3r, 3r^2, 3r^3=81$. Then $r^3=27$, so $r=3$. The two inserted numbers are $3r=9$ and $3r^2=27$.

For the expression to be the geometric mean, $\left(\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}\right)^2=ab$. Thus $(a^{n+1}+b^{n+1})^2=ab(a^n+b^n)^2$. After cancelling the common middle term, $a^{2n+2}+b^{2n+2}=a^{2n+1}b+ab^{2n+1}$. This gives $(a-b)(a^{2n+1}-b^{2n+1})=0$. For arbitrary $a$ and $b$, take $2n+1=0$, so $n=-\dfrac12$.

Let the positive numbers be $x$ and $y$. Given $x+y=6\sqrt{xy}$. Put $u=\sqrt{x/y}$. Dividing by $\sqrt{xy}$ gives $u+\dfrac1u=6$, so $u^2-6u+1=0$. Hence $u=3\pm2\sqrt2$. Therefore $x:y=u^2:1$. Since $(3+2\sqrt2)(3-2\sqrt2)=1$, this ratio is $(3+2\sqrt2):(3-2\sqrt2)$.

Let the two positive numbers be $x$ and $y$. Then $\dfrac{x+y}{2}=A$ and $\sqrt{xy}=G$, so $x+y=2A$ and $xy=G^2$. The numbers are roots of $t^2-2At+G^2=0$. Solving, $t=A\pm\sqrt{A^2-G^2}=A\pm\sqrt{(A+G)(A-G)}$.

The population doubles each hour. Starting with 30, after $k$ hours the population is $30\cdot2^k$. Thus after 2 hours it is $30\cdot2^2=120$, after 4 hours it is $30\cdot2^4=480$, and after $n$ hours it is $30\cdot2^n$.

For annual compound interest, $A=P\left(1+\dfrac{r}{100}\right)^n$. Here $P=500$, $r=10$ and $n=10$. Thus $A=500(1.1)^{10}=500\left(\dfrac{11}{10}\right)^{10}=1296.87123\ldots$. Therefore the amount is approximately Rs 1296.87.

Let the roots be $\alpha$ and $\beta$. Since their A.M. is 8, $\dfrac{\alpha+\beta}{2}=8$, so $\alpha+\beta=16$. Since their G.M. is 5, $\sqrt{\alpha\beta}=5$, so $\alpha\beta=25$. The quadratic equation is $x^2-(\alpha+\beta)x+\alpha\beta=0$, i.e. $x^2-16x+25=0$.

Since $f(x+y)=f(x)f(y)$ and $f(1)=3$, we get $f(2)=9$, $f(3)=27$, and in general $f(x)=3^x$. Thus $3+3^2+\cdots+3^n=120$. Using the G.P. sum, $\dfrac{3(3^n-1)}{3-1}=120$, so $3^n-1=80$ and $3^n=81=3^4$. Hence $n=4$.

For the G.P., $a=5$ and $r=2$. The sum of $n$ terms is $S_n=\dfrac{5(2^n-1)}{2-1}=5(2^n-1)$. Given $5(2^n-1)=315$, so $2^n-1=63$ and $2^n=64=2^6$. Hence $n=6$. The last term is $a_6=5\cdot2^5=160$.

Let the common ratio be $r$. Since the first term is 1, the third and fifth terms are $r^2$ and $r^4$. Given $r^2+r^4=90$. Put $u=r^2$. Then $u^2+u-90=0$, so $(u-9)(u+10)=0$. Since $u=r^2\ge0$, $u=9$, giving $r=\pm3$.

Let the three G.P. numbers be $x,y,z$. Then $x+y+z=56$ and $y^2=xz$. Since $x-1,y-7,z-21$ are in A.P., $2(y-7)=(x-1)+(z-21)$, so $x+z-2y=8$. Using $x+z=56-y$, we get $56-y-2y=8$, hence $y=16$ and $x+z=40$. Also $xz=16^2=256$. Thus $x$ and $z$ are roots of $t^2-40t+256=0$, giving $t=8,32$. The numbers are $8,16,32$ or $32,16,8$.

Let the G.P. have $2m$ terms, first term $a$ and common ratio $r$. The sum of odd-place terms is $a+ar^2+\cdots+ar^{2m-2}=\dfrac{a(1-r^{2m})}{1-r^2}$. The sum of all terms is $\dfrac{a(1-r^{2m})}{1-r}$. Therefore $\dfrac{\text{sum of all terms}}{\text{sum of odd-place terms}}=\dfrac{1-r^2}{1-r}=1+r$. Given this ratio is 5, $1+r=5$, so $r=4$.

From $\dfrac{a+bx}{a-bx}=\dfrac{b+cx}{b-cx}$, cross-multiplication gives $(a+bx)(b-cx)=(b+cx)(a-bx)$. Expanding and simplifying gives $2x(b^2-ac)=0$. Since $x\ne0$, $b^2=ac$. Similarly, comparing the second and third fractions gives $c^2=bd$. Hence $\dfrac ba=\dfrac cb=\dfrac dc$ where defined, so $a,b,c,d$ are in G.P.

Let the $n$ terms be $a,ar,\ldots,ar^{n-1}$. Then $S=a(1+r+\cdots+r^{n-1})$. Also $P=a^nr^{n(n-1)/2}$ and $R=\dfrac1a(1+r^{-1}+\cdots+r^{-(n-1)})=\dfrac1a r^{-(n-1)}(1+r+\cdots+r^{n-1})$. Hence $P^2R^n=a^{2n}r^{n(n-1)}\cdot a^{-n}r^{-n(n-1)}(1+r+\cdots+r^{n-1})^n=a^n(1+r+\cdots+r^{n-1})^n=S^n$.

Let $a,b,c,d$ be $a,ar,ar^2,ar^3$. Then $a^n+b^n=a^n(1+r^n)$, $b^n+c^n=a^nr^n(1+r^n)$, and $c^n+d^n=a^nr^{2n}(1+r^n)$. These three expressions have common ratio $r^n$, so they are in G.P.

Let $a,b,c,d$ be $A,Ar,Ar^2,Ar^3$. Since $a$ and $b$ are roots of $x^2-3x+p=0$, $A(1+r)=3$ and $p=A^2r$. Since $c$ and $d$ are roots of $x^2-12x+q=0$, $Ar^2(1+r)=12$ and $q=A^2r^5$. Dividing the sums gives $r^2=4$, so $q/p=r^4=16$. Hence $q=16p$, and $(q+p):(q-p)=17p:15p=17:15$.

Let the A.M. be $A$ and the G.M. be $G$. Since $A:G=m:n$, write $A=km$ and $G=kn$. From the standard result for two positive numbers with A.M. $A$ and G.M. $G$, the numbers are $A\pm\sqrt{A^2-G^2}$. Therefore they are $km\pm k\sqrt{m^2-n^2}$. Their ratio is $(m+\sqrt{m^2-n^2}):(m-\sqrt{m^2-n^2})$.

(i) The $k^{\text{th}}$ term is $\dfrac59(10^k-1)$. Thus the sum is $\dfrac59\left(\dfrac{10(10^n-1)}9-n\right)=\dfrac{5(10^{n+1}-10-9n)}{81}$. (ii) The $k^{\text{th}}$ term is $0.\underbrace{66\ldots6}_{k}=\dfrac23(1-10^{-k})$. Therefore the sum is $\dfrac23\left(n-\sum_{k=1}^n10^{-k}\right)=\dfrac{2n}{3}-\dfrac{2}{27}(1-10^{-n})$.

The $k^{\text{th}}$ term is $(2k)(2k+2)=4k(k+1)$. Therefore the $20^{\text{th}}$ term is $4\cdot20\cdot21=1680$.

The unpaid balance is Rs 6000, paid in annual principal instalments of Rs 500, so there are 12 instalments. Interest is charged on unpaid amounts Rs 6000, Rs 5500, ..., Rs 500. Total interest is $12\%$ of this sum: $0.12\times(6000+5500+\cdots+500)=0.12\times500(12+11+\cdots+1)=60\times78=4680$. Total cost is $12000+4680=16680$ rupees.

The unpaid balance is Rs 18000, paid in annual principal instalments of Rs 1000, so there are 18 instalments. Interest is charged on unpaid amounts Rs 18000, Rs 17000, ..., Rs 1000. Total interest is $10\%$ of this sum: $0.10\times1000(18+17+\cdots+1)=100\times171=17100$. Total cost is $22000+17100=39100$ rupees.

The number of letters in the first set is $4$, and each set has 4 times as many letters as the previous set. Hence the number of letters in the $8^{\text{th}}$ set is $4^8=65536$. At 50 paise, or Rs $0.50$, per letter, the postage is $65536\times0.50=32768$ rupees.

Simple interest per year is $\dfrac{10000\times5}{100}=500$ rupees. The amount after $t$ years is $10000+500t$. Therefore the amount in the $15^{\text{th}}$ year is $10000+500\times15=17500$ rupees. After 20 years, the amount is $10000+500\times20=20000$ rupees.

Each year the machine retains $80\%$ of its value. After 5 years, the value is $15625(0.8)^5=15625\left(\dfrac45\right)^5=15625\cdot\dfrac{1024}{3125}=5120$ rupees.

Let the job originally require $x$ days with 150 workers, so total work is $150x$ worker-days. With dropouts, the number of workers over $x+8$ days is $150,146,142,\ldots$. Hence actual work done is $\dfrac{x+8}{2}\{2(150)+(x+8-1)(-4)\}=(x+8)(136-2x)$. Set this equal to $150x$: $(x+8)(136-2x)=150x$. This simplifies to $x^2-60x-544=0$, giving $x=68$ as the positive root. Therefore the work was completed in $x+8=76$ days.