Class 11 Maths NCERT Solutions Chapter 8 Miscellaneous Exercise – Sequences and Series

Since f(x+y)=f(x)f(y) and f(1)=3, setting x=y=1 gives f(2)=f(1)f(1)=9=3^2. In general, f(n)=f(1)^n=3^n, so f(1),f(2),f(3),\ldots is a G.P. with first term 3 and common ratio 3.

The given sum condition becomes:

\sum_{x=1}^{n}3^x=\dfrac{3(3^n-1)}{3-1}=120

3(3^n-1)=240 \ \Rightarrow\ 3^n-1=80 \ \Rightarrow\ 3^n=81=3^4

Here a=5, r=2. Setting S_n=315:

\dfrac{5(2^n-1)}{2-1}=315 \ \Rightarrow\ 2^n-1=63 \ \Rightarrow\ 2^n=64=2^6

n=6

The last (6th) term is:

a_6=ar^5=5(2)^5=5\times32=160

With a=1, the third and fifth terms are r^2 and r^4. The condition gives:

r^2+r^4=90

Letting u=r^2:

u^2+u-90=0 \ \Rightarrow\ (u-9)(u+10)=0 \ \Rightarrow\ u=9 \text{ or } u=-10

Since u=r^2\ge0, reject u=-10. So r^2=9, giving r=\pm3.

Let the three numbers in G.P. be a,\ ar,\ ar^2. The sum condition gives:

a+ar+ar^2=56 \qquad \ldots(1)

After subtracting 1, 7, 21, the numbers a-1,\ ar-7,\ ar^2-21 form an A.P., so twice the middle term equals the sum of the outer two:

2(ar-7)=(a-1)+(ar^2-21) \qquad \ldots(2)

Solving (1) and (2) together gives two possible solutions: (a,r)=(8,2) or (a,r)=(32,\tfrac12) — the same three numbers listed in opposite order.

Taking a=8,\ r=2, the numbers are:

8,\ 8(2)=16,\ 8(4)=32

Check: sum =8+16+32=56 ✓. Subtracting 1, 7, 21 gives 7,\ 9,\ 11, which is indeed an A.P. with common difference 2.

Let the G.P. have 2m terms, first term a, common ratio r. The sum of all terms is:

S_{\text{all}}=\dfrac{a(r^{2m}-1)}{r-1}

The terms in odd places (1st, 3rd, 5th, ..., (2m-1)th) themselves form a G.P. with first term a and common ratio r^2, with m terms:

S_{\text{odd}}=\dfrac{a\left((r^2)^m-1\right)}{r^2-1}=\dfrac{a(r^{2m}-1)}{r^2-1}

The given condition S_{\text{all}}=5\,S_{\text{odd}} becomes:

\dfrac{a(r^{2m}-1)}{r-1}=5\cdot\dfrac{a(r^{2m}-1)}{r^2-1}=\dfrac{5a(r^{2m}-1)}{(r-1)(r+1)}

Cancelling the common non-zero factors:

1=\dfrac{5}{r+1} \ \Rightarrow\ r+1=5 \ \Rightarrow\ r=4

Taking the first equality and cross-multiplying:

(a+bx)(b-cx)=(a-bx)(b+cx)

Expanding both sides:

ab-acx+b^2x-bcx^2=ab+acx-b^2x-bcx^2

Cancelling ab and -bcx^2 from both sides, and collecting like terms:

-acx+b^2x=acx-b^2x \ \Rightarrow\ 2b^2x=2acx

Since x\ne0, dividing by 2x gives b^2=ac.

By exactly the same argument applied to the second equality, c^2=bd.

From b^2=ac: \dfrac{b}{a}=\dfrac{c}{b}. From c^2=bd: \dfrac{c}{b}=\dfrac{d}{c}. So:

\dfrac{b}{a}=\dfrac{c}{b}=\dfrac{d}{c}

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

S=\dfrac{a(r^n-1)}{r-1} \qquad P=a^n\,r^{\frac{n(n-1)}{2}}

The reciprocals \dfrac{1}{a},\dfrac{1}{ar},\ldots,\dfrac{1}{ar^{n-1}} themselves form a G.P. with first term \dfrac1a and common ratio \dfrac1r, so:

R=\dfrac{\frac1a\left(1-\left(\frac1r\right)^n\right)}{1-\frac1r}=\dfrac{r^{n-1}(r^n-1)}{a\,r^n(r-1)}=\dfrac{r^n-1}{a\,r^{n-1}(r-1)}

Now computing P^2R^n:

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

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

P^2R^n=a^{2n}r^{n(n-1)}\cdot\dfrac{(r^n-1)^n}{a^n r^{n(n-1)}(r-1)^n}=a^n\cdot\dfrac{(r^n-1)^n}{(r-1)^n}=\left(\dfrac{a(r^n-1)}{r-1}\right)^n=S^n

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

The three expressions become:

T_1=a^n+b^n=a^n(1+r^n)

T_2=b^n+c^n=a^nr^n(1+r^n)

T_3=c^n+d^n=a^nr^{2n}(1+r^n)

Checking whether T_2^2=T_1T_3:

T_2^2=a^{2n}r^{2n}(1+r^n)^2

T_1T_3=a^n(1+r^n)\cdot a^nr^{2n}(1+r^n)=a^{2n}r^{2n}(1+r^n)^2

Both sides are equal.

By the sum and product of roots: a+b=3,\ ab=p and c+d=12,\ cd=q.

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

a+ar=3 \ \Rightarrow\ a(1+r)=3 \qquad \ldots(1)

ar^2+ar^3=12 \ \Rightarrow\ ar^2(1+r)=12 \qquad \ldots(2)

Dividing (2) by (1):

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

Taking r=2: from (1), a(3)=3\Rightarrow a=1. Then:

p=ab=a(ar)=a^2r=1\times2=2 \qquad q=cd=(ar^2)(ar^3)=a^2r^5=1\times32=32

\dfrac{q+p}{q-p}=\dfrac{32+2}{32-2}=\dfrac{34}{30}=\dfrac{17}{15}

(Taking r=-2 instead gives a=-3, p=-18, q=-288, and the same ratio 17:15.)

Given \dfrac{\text{A.M.}}{\text{G.M.}}=\dfrac{m}{n}, i.e. \dfrac{(a+b)/2}{\sqrt{ab}}=\dfrac{m}{n}, so n(a+b)=2m\sqrt{ab}.

Let t=\sqrt{\dfrac{a}{b}}, so a=bt^2 and \sqrt{ab}=bt. Substituting:

n\,b(t^2+1)=2m\,bt \ \Rightarrow\ n(t^2+1)=2mt \ \Rightarrow\ nt^2-2mt+n=0

Solving this quadratic in t:

t=\dfrac{2m\pm\sqrt{4m^2-4n^2}}{2n}=\dfrac{m\pm\sqrt{m^2-n^2}}{n}

Taking the positive root t=\dfrac{m+\sqrt{m^2-n^2}}{n}, and using \dfrac{a}{b}=t^2:

\dfrac{a}{b}=\dfrac{\left(m+\sqrt{m^2-n^2}\right)^2}{n^2}

Since \left(m+\sqrt{m^2-n^2}\right)\left(m-\sqrt{m^2-n^2}\right)=m^2-(m^2-n^2)=n^2, dividing the numerator and denominator of the ratio \left(m+\sqrt{m^2-n^2}\right):\left(m-\sqrt{m^2-n^2}\right) by n^2 shows it equals \dfrac{\left(m+\sqrt{m^2-n^2}\right)^2}{n^2}, which is exactly \dfrac{a}{b} found above.

Factor out 5 and rewrite each term as a power of 10 minus 1:

S_n=5\left[1+11+111+\cdots\right]=\dfrac{5}{9}\left[9+99+999+\cdots\right]

=\dfrac{5}{9}\left[(10-1)+(10^2-1)+\cdots+(10^n-1)\right]=\dfrac{5}{9}\left[(10+10^2+\cdots+10^n)-n\right]

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

S_n=\dfrac{5}{9}\left[\dfrac{10(10^n-1)}{9}-n\right]

Factor out 6 and rewrite each term similarly:

S_n=6\left[0.1+0.11+0.111+\cdots\right]=\dfrac{6}{9}\left[0.9+0.99+0.999+\cdots\right]

=\dfrac{2}{3}\left[(1-0.1)+(1-0.01)+\cdots+(1-0.1^n)\right]=\dfrac{2}{3}\left[n-(0.1+0.01+\cdots+0.1^n)\right]

The bracketed sum inside is a G.P. with a=0.1,\ r=0.1:

0.1+0.01+\cdots+0.1^n=\dfrac{0.1(1-0.1^n)}{1-0.1}=\dfrac{1}{9}\left(1-0.1^n\right)

So:

S_n=\dfrac{2}{3}\left[n-\dfrac{1}{9}\left(1-0.1^n\right)\right]

The kth term of this series is the product of the kth terms of 2, 4, 6, ... and 4, 6, 8, ..., which are 2k and 2k+2 respectively:

T_k=2k(2k+2)=4k(k+1)

For the 20th term, k=20:

T_{20}=4(20)(21)=1680

The unpaid balance after the cash payment is 12000-6000=6000, repaid in fixed instalments of Rs 500, so the number of instalments is \dfrac{6000}{500}=12.

The unpaid amount before each instalment is 6000, 5500, 5000, ..., 500 — an A.P. with 12 terms, first term 6000, last term 500. The total interest paid is 12% of the sum of these unpaid amounts:

\text{Sum of unpaid amounts}=\dfrac{12}{2}(6000+500)=6\times6500=39000

\text{Total interest}=12\%\text{ of }39000=0.12\times39000=4680

The total cost is the cash paid, plus the total of the principal instalments (which equals the original balance of 6000), plus the total interest:

\text{Total cost}=6000+6000+4680=16680

The unpaid balance after the cash payment is 22000-4000=18000, repaid in instalments of Rs 1000, so the number of instalments is \dfrac{18000}{1000}=18.

The unpaid amount before each instalment is 18000, 17000, ..., 1000 — an A.P. with 18 terms, first term 18000, last term 1000:

\text{Sum of unpaid amounts}=\dfrac{18}{2}(18000+1000)=9\times19000=171000

\text{Total interest}=10\%\text{ of }171000=0.10\times171000=17100

The total cost is:

\text{Total cost}=4000+18000+17100=39100

The number of letters mailed in each set forms a G.P.: 4 letters in the 1st set, 4^2=16 in the 2nd, and so on, with 4^k letters in the kth set.

The total number of letters mailed by the time the 8th set is mailed is the sum of this G.P. up to 8 terms, with a=4,\ r=4:

S_8=\dfrac{4(4^8-1)}{4-1}=\dfrac{4(65536-1)}{3}=\dfrac{4\times65535}{3}=87380

At 50 paise (Rs 0.50) per letter, the total postage cost is:

87380\times0.50=43690

With simple interest, the same interest amount is earned every year:

\text{Annual interest}=5\%\text{ of }10000=500

Since this interest is fixed each year, the total accumulated amount after k years, 10000+500k, forms an arithmetic sequence in k.

The amount at the start of the 15th year (i.e. after 14 complete years of interest) is:

10000+500(14)=10000+7000=17000

The total amount after 20 complete years is:

10000+500(20)=10000+10000=20000

Depreciating by 20% each year means the value is multiplied by 1-\dfrac{20}{100}=\dfrac{4}{5} every year, so the value after n years is a G.P. term:

\text{Value after }n\text{ years}=15625\left(\dfrac{4}{5}\right)^n

For n=5:

15625\left(\dfrac{4}{5}\right)^5=15625\times\dfrac{1024}{3125}=5\times1024=5120

Let the job have originally been planned to finish in n days with all 150 workers throughout, so the total work is 150n worker-days.

With the drop-outs, the number of workers each day forms an A.P.: 150 on day 1, 146 on day 2, 142 on day 3, and so on, decreasing by 4 each day, over n+8 actual days (since it took 8 more days than planned).

The total work done equals the sum of this A.P. over n+8 days, which must still equal the original total work 150n:

\dfrac{(n+8)}{2}\Big[2(150)+(n+8-1)(-4)\Big]=150n

Simplifying inside the bracket:

\dfrac{(n+8)}{2}\Big[300-4(n+7)\Big]=150n

(n+8)(300-4n-28)=300n

(n+8)(272-4n)=300n

Expanding the left-hand side:

272n-4n^2+2176-32n=300n \ \Rightarrow\ -4n^2+240n+2176=300n

Rearranging into a standard quadratic and dividing by -4:

n^2+15n-544=0

This factors as (n-17)(n+32)=0, giving n=17 or n=-32. Since n must be a positive number of days, n=17.

So the work was originally planned for 17 days, and actually took 17+8=25 days.