Class 11 Maths NCERT Solutions Chapter 8 Ex 8.2 – Sequences and Series

Here a=\dfrac{5}{2} and r=\dfrac{5/4}{5/2}=\dfrac{1}{2}.

a_{20}=ar^{19}=\dfrac{5}{2}\left(\dfrac{1}{2}\right)^{19}=\dfrac{5}{2^{20}}

a_n=ar^{n-1}=\dfrac{5}{2}\left(\dfrac{1}{2}\right)^{n-1}=\dfrac{5}{2^{n}}

Given a_8=ar^7=192 and r=2, so a=\dfrac{192}{2^7}=\dfrac{192}{128}=\dfrac{3}{2}.

a_{12}=ar^{11}=\dfrac{3}{2}(2)^{11}=\dfrac{3}{2}\times2048=3072

Let the G.P. have first term a and common ratio r. Then:

p=ar^4 \qquad q=ar^7 \qquad s=ar^{10}

Computing q^2 and ps separately:

q^2=(ar^7)^2=a^2r^{14}

ps=(ar^4)(ar^{10})=a^2r^{14}

Since both equal a^2r^{14}, q^2=ps.

Let the common ratio be r, with a=-3. The condition a_4=(a_2)^2 gives:

ar^3=(ar)^2=a^2r^2

Since a\ne0, dividing both sides by ar^2:

r=a=-3

So the 7th term is:

a_7=ar^6=(-3)(-3)^6=(-3)(729)=-2187

Here a=2, r=\dfrac{2\sqrt2}{2}=\sqrt2. Setting a_n=128:

2(\sqrt2)^{n-1}=128 \ \Rightarrow\ 2^{\frac{n-1}{2}}=2^{6} \ \Rightarrow\ \dfrac{n-1}{2}=6 \ \Rightarrow\ n=13

Here a=\sqrt3, r=\dfrac{3}{\sqrt3}=\sqrt3. Setting a_n=729:

\sqrt3\,(\sqrt3)^{n-1}=(\sqrt3)^n=729=3^6=(\sqrt3)^{12} \ \Rightarrow\ n=12

Here a=\dfrac{1}{3}, r=\dfrac{1}{3}. Setting a_n=\dfrac{1}{19683}:

\left(\dfrac{1}{3}\right)^n=\dfrac{1}{19683}=\dfrac{1}{3^9} \ \Rightarrow\ n=9

Three numbers are in G.P. exactly when the square of the middle term equals the product of the outer two:

x^2=\left(-\dfrac{2}{7}\right)\left(-\dfrac{7}{2}\right)=1

x=\pm1

Here a=0.15, r=0.1, n=20. Since r\ne1:

S_{20}=\dfrac{a(1-r^{20})}{1-r}=\dfrac{0.15\left(1-(0.1)^{20}\right)}{0.9}=\dfrac{1}{6}\left(1-(0.1)^{20}\right)

Here a=\sqrt7 and r=\dfrac{\sqrt{21}}{\sqrt7}=\sqrt3. Since r\ne1:

S_n=\dfrac{a(r^n-1)}{r-1}=\dfrac{\sqrt7\left((\sqrt3)^n-1\right)}{\sqrt3-1}

Here a_1=1 and the common ratio is -a. Since a\ne-1, the common ratio is not 1, so:

S_n=\dfrac{1\cdot\left(1-(-a)^n\right)}{1-(-a)}=\dfrac{1-(-a)^n}{1+a}

Here a=x^3 and r=\dfrac{x^5}{x^3}=x^2. Since x\ne\pm1, r\ne1, so:

S_n=\dfrac{x^3\left((x^2)^n-1\right)}{x^2-1}=\dfrac{x^3(x^{2n}-1)}{x^2-1}

Splitting the sum into a constant part and a G.P. part:

\sum_{k=1}^{11}(2+3^k)=\sum_{k=1}^{11}2+\sum_{k=1}^{11}3^k=(11\times2)+\sum_{k=1}^{11}3^k

The second sum is a G.P. with a=3, r=3, n=11:

\sum_{k=1}^{11}3^k=\dfrac{3(3^{11}-1)}{3-1}=\dfrac{3(177147-1)}{2}=\dfrac{3\times177146}{2}=265719

Adding the constant part:

22+265719=265741

Let the three terms be \dfrac{a}{r},\ a,\ ar. Then:

\dfrac{a}{r}+a+ar=\dfrac{39}{10} \qquad \left(\dfrac{a}{r}\right)(a)(ar)=1

The second equation gives a^3=1, so a=1 (taking the real root).

Substituting a=1 into the first equation:

\dfrac{1}{r}+1+r=\dfrac{39}{10} \ \Rightarrow\ 10+10r+10r^2=39r \ \Rightarrow\ 10r^2-29r+10=0

Solving this quadratic in r:

r=\dfrac{29\pm\sqrt{841-400}}{20}=\dfrac{29\pm21}{20}

r=\dfrac{5}{2} \text{ or } r=\dfrac{2}{5}

For r=\dfrac{5}{2}: terms are \dfrac{2}{5},\,1,\,\dfrac{5}{2}. For r=\dfrac{2}{5}: terms are \dfrac{5}{2},\,1,\,\dfrac{2}{5} — the same three numbers in reverse order.

Here a=3, r=3. Setting S_n=120:

\dfrac{3(3^n-1)}{3-1}=120 \ \Rightarrow\ 3(3^n-1)=240 \ \Rightarrow\ 3^n-1=80 \ \Rightarrow\ 3^n=81=3^4

n=4

Let the first term be a and common ratio r. The first three terms sum to:

a(1+r+r^2)=16 \qquad \ldots(1)

The next three terms (4th, 5th, 6th) sum to:

ar^3(1+r+r^2)=128 \qquad \ldots(2)

Dividing (2) by (1):

r^3=\dfrac{128}{16}=8 \ \Rightarrow\ r=2

Substituting r=2 into (1):

a(1+2+4)=16 \ \Rightarrow\ 7a=16 \ \Rightarrow\ a=\dfrac{16}{7}

The sum to n terms is:

S_n=\dfrac{a(r^n-1)}{r-1}=\dfrac{\frac{16}{7}(2^n-1)}{1}=\dfrac{16(2^n-1)}{7}

Using a_7=ar^6=64 with a=729:

r^6=\dfrac{64}{729} \ \Rightarrow\ r=\sqrt[6]{\dfrac{64}{729}}=\dfrac{2}{3}

Then:

S_7=\dfrac{a(1-r^7)}{1-r}=\dfrac{729\left(1-\left(\frac{2}{3}\right)^7\right)}{1-\frac{2}{3}}=2187\left(1-\dfrac{128}{2187}\right)=2187-128

Let the first term be a and common ratio r. The condition a_5=4a_3 gives:

ar^4=4ar^2 \ \Rightarrow\ r^2=4 \ \Rightarrow\ r=\pm2

The condition a+ar=-4 gives a(1+r)=-4.

For r=2: a(3)=-4 \Rightarrow a=-\dfrac{4}{3}, giving the G.P. -\dfrac{4}{3},-\dfrac{8}{3},-\dfrac{16}{3},\ldots

For r=-2: a(-1)=-4 \Rightarrow a=4, giving the G.P. 4,-8,16,-32,\ldots

Let the G.P. have first term A and common ratio R. Then:

x=AR^3 \qquad y=AR^9 \qquad z=AR^{15}

Computing y^2 and xz:

y^2=(AR^9)^2=A^2R^{18} \qquad xz=(AR^3)(AR^{15})=A^2R^{18}

Since y^2=xz, the numbers x, y, z satisfy the defining condition for three terms in G.P.

This sequence is not itself a G.P., but each term can be related to one by factoring out 8 and rewriting each term as a power of 10 minus 1:

S_n=8+88+888+\cdots \text{ to } n \text{ terms}

=\dfrac{8}{9}\left[9+99+999+\cdots \text{ to } n \text{ terms}\right]

=\dfrac{8}{9}\left[(10-1)+(10^2-1)+(10^3-1)+\cdots \text{ to } n \text{ terms}\right]

=\dfrac{8}{9}\left[(10+10^2+\cdots+10^n)-n\right]

The bracketed sum is now a genuine G.P. with a=10, r=10:

=\dfrac{8}{9}\left[\dfrac{10(10^n-1)}{9}-n\right]

Multiplying the corresponding terms:

2\times128=256 \quad 4\times32=128 \quad 8\times8=64 \quad 16\times2=32 \quad 32\times\dfrac{1}{2}=16

These products, 256, 128, 64, 32, 16, form a G.P. with a=256, r=\dfrac{1}{2}, n=5:

S_5=\dfrac{256\left(1-\left(\frac12\right)^5\right)}{1-\frac12}=512\left(1-\dfrac{1}{32}\right)=512\times\dfrac{31}{32}=496

The kth term of the first sequence is ar^{k-1}, and the kth term of the second is AR^{k-1}. Their product is:

T_k=(ar^{k-1})(AR^{k-1})=aA(rR)^{k-1}

This is exactly the form of the kth term of a G.P. with first term aA and common ratio rR, since the ratio of consecutive terms is:

\dfrac{T_{k+1}}{T_k}=\dfrac{aA(rR)^k}{aA(rR)^{k-1}}=rR \quad\text{(constant)}

Let the four numbers be a,\ ar,\ ar^2,\ ar^3. The conditions give:

ar^2-a=9 \qquad \ldots(1)

ar-ar^3=18 \qquad \ldots(2)

From (1): a(r^2-1)=9. From (2): ar(1-r^2)=18, i.e. -ar(r^2-1)=18.

Dividing the second by the first:

\dfrac{-ar(r^2-1)}{a(r^2-1)}=\dfrac{18}{9} \ \Rightarrow\ -r=2 \ \Rightarrow\ r=-2

Substituting r=-2 into (1):

a(4-1)=9 \ \Rightarrow\ 3a=9 \ \Rightarrow\ a=3

So the four numbers are:

3,\ 3(-2)=-6,\ 3(4)=12,\ 3(-8)=-24

Let the G.P. have first term A and common ratio R. Then:

a=AR^{p-1} \qquad b=AR^{q-1} \qquad c=AR^{r-1}

Substituting into the left-hand side:

a^{q-r}b^{r-p}c^{p-q}=\left(AR^{p-1}\right)^{q-r}\left(AR^{q-1}\right)^{r-p}\left(AR^{r-1}\right)^{p-q}

The powers of A add up to (q-r)+(r-p)+(p-q)=0, so the A factor becomes A^0=1.

The powers of R add up to (p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q), which expands and simplifies to 0 as well, so the R factor is also R^0=1.

Let the common ratio be r, so b=ar^{n-1}. The product of the n terms is:

P=a\cdot ar\cdot ar^2\cdots ar^{n-1}=a^n\,r^{0+1+2+\cdots+(n-1)}=a^n\,r^{\frac{n(n-1)}{2}}

Squaring:

P^2=a^{2n}\,r^{n(n-1)}

On the other hand:

(ab)^n=\left(a\cdot ar^{n-1}\right)^n=\left(a^2r^{n-1}\right)^n=a^{2n}r^{n(n-1)}

Both sides equal a^{2n}r^{n(n-1)}.

Let the G.P. have first term a and common ratio r\ne1. Then:

S_n=\dfrac{a(r^n-1)}{r-1}

The sum of terms from the (n+1)th to the (2n)th is S_{2n}-S_n:

S_{2n}-S_n=\dfrac{a(r^{2n}-1)}{r-1}-\dfrac{a(r^n-1)}{r-1}=\dfrac{a\left(r^{2n}-r^n\right)}{r-1}=\dfrac{a\,r^n(r^n-1)}{r-1}=r^n\,S_n

Therefore, the required ratio is:

\dfrac{S_n}{S_{2n}-S_n}=\dfrac{S_n}{r^n\,S_n}=\dfrac{1}{r^n}

Since a, b, c, d are in G.P., write b=ar,\ c=ar^2,\ d=ar^3 for common ratio r.

The left-hand side becomes:

(a^2+a^2r^2+a^2r^4)(a^2r^2+a^2r^4+a^2r^6)=a^2(1+r^2+r^4)\cdot a^2r^2(1+r^2+r^4)

=a^4r^2(1+r^2+r^4)^2

The right-hand side becomes:

(a\cdot ar+ar\cdot ar^2+ar^2\cdot ar^3)^2=\left(a^2r+a^2r^3+a^2r^5\right)^2=\left(a^2r(1+r^2+r^4)\right)^2

=a^4r^2(1+r^2+r^4)^2

Both sides simplify to the same expression.

Let G_1, G_2 be the two numbers, so that 3, G_1, G_2, 81 is a 4-term G.P. Then 81 is the 4th term:

81=3\cdot r^3 \ \Rightarrow\ r^3=27 \ \Rightarrow\ r=3 (taking the real root)

G_1=3r=9 \qquad G_2=3r^2=27

The geometric mean of a and b is \sqrt{ab}, so the condition is:

\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{ab}=a^{1/2}b^{1/2}

Try n=-\dfrac12. Then the left-hand side becomes:

\dfrac{a^{1/2}+b^{1/2}}{a^{-1/2}+b^{-1/2}}=\dfrac{a^{1/2}+b^{1/2}}{\dfrac{1}{a^{1/2}}+\dfrac{1}{b^{1/2}}}=\dfrac{a^{1/2}+b^{1/2}}{\dfrac{a^{1/2}+b^{1/2}}{a^{1/2}b^{1/2}}}=a^{1/2}b^{1/2}=\sqrt{ab}

This confirms the condition holds exactly when n=-\dfrac12.

Let the numbers be a and b. The given condition is a+b=6\sqrt{ab}. Dividing both sides by \sqrt{ab}:

\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}=6

Let y=\sqrt{\dfrac{a}{b}}. Then y+\dfrac{1}{y}=6, i.e. y^2-6y+1=0.

Solving this quadratic:

y=\dfrac{6\pm\sqrt{36-4}}{2}=\dfrac{6\pm\sqrt{32}}{2}=3\pm2\sqrt2

Taking y=3+2\sqrt2 (the case y=3-2\sqrt2 just swaps the roles of a and b), and using y^2=\dfrac{a}{b}:

\dfrac{a}{b}=(3+2\sqrt2)^2

Since (3+2\sqrt2)(3-2\sqrt2)=9-8=1, dividing numerator and denominator of the ratio (3+2\sqrt2):(3-2\sqrt2) shows it equals (3+2\sqrt2)^2:1, matching \dfrac{a}{b} exactly.

Let the two positive numbers be a and b. By definition:

A=\dfrac{a+b}{2} \ \Rightarrow\ a+b=2A \qquad G=\sqrt{ab} \ \Rightarrow\ ab=G^2

Since a and b are the roots of the quadratic t^2-(a+b)t+ab=0, they satisfy:

t^2-2At+G^2=0

Solving using the quadratic formula:

t=\dfrac{2A\pm\sqrt{4A^2-4G^2}}{2}=A\pm\sqrt{A^2-G^2}

Since A^2-G^2=(A-G)(A+G):

t=A\pm\sqrt{(A-G)(A+G)}

The bacteria count forms a G.P. with first term a=30 (the initial count) and common ratio r=2 (doubling every hour). After k hours, the count is 30\times2^k.

End of 2nd hour: 30\times2^2=30\times4=120

End of 4th hour: 30\times2^4=30\times16=480

End of nth hour: 30\times2^n

With compound interest, the amount after each year is the previous amount multiplied by 1+\dfrac{10}{100}=1.1. After n years, the amount is:

\text{Amount}=500(1.1)^n

For n=10:

\text{Amount}=500(1.1)^{10}

Let the roots be \alpha and \beta. Given \text{A.M.}=\dfrac{\alpha+\beta}{2}=8 and \text{G.M.}=\sqrt{\alpha\beta}=5:

\alpha+\beta=16 \qquad \alpha\beta=25

A quadratic with roots \alpha,\beta is x^2-(\alpha+\beta)x+\alpha\beta=0:

x^2-16x+25=0