Samacheer Class 10 Maths - Numbers and Sequences

By Euclid’s Division Lemma,

[ n=3q+r ]

where (0\le r<3).

Given remainder:

[ r=2 ]

Hence,

[ n=3q+2 ]

for positive integers (q=0,1,2,3,\dots)

Thus the numbers are:

[ 2,5,8,11,14,\dots ]

### Answer

[ 2,5,8,11,\dots ]

---

Divide 532 by 21.

[ 532=21\times25+7 ]

Thus,

Quotient (=25) Remainder (=7)

### Answer

Completed rows:

[ 25 ]

Flower pots left over:

[ 7 ]

---

### Proof

Let two consecutive positive integers be:

[ n \text{ and } n+1 ]

Their product is:

[ n(n+1) ]

Among any two consecutive integers, one must be even.

Therefore the product contains a factor 2.

Hence,

[ n(n+1) ]

is divisible by 2.

### Proved.

---

Given:

[ a\equiv9 \pmod{13} ]

[ b\equiv7 \pmod{13} ]

[ c\equiv10 \pmod{13} ]

Adding,

[ a+b+c\equiv9+7+10 ]

[ \equiv26 ]

[ \equiv0\pmod{13} ]

Hence,

[ a+b+c ]

is divisible by 13.

### Proved.

---

### Proof

Any integer is either:

[ 2n \quad \text{or} \quad 2n+1 ]

### Case 1: Even integer

[ (2n)^2=4n^2 ]

which leaves remainder 0 when divided by 4.

### Case 2: Odd integer

[ (2n+1)^2 ]

[ =4n^2+4n+1 ]

[ =4n(n+1)+1 ]

which leaves remainder 1 when divided by 4.

Hence the square of any integer leaves remainder either 0 or 1 when divided by 4.

### Proved.

---

---

Subtract the remainder from each number:

[ 1230-12=1218 ]

[ 1926-12=1914 ]

Required number:

[ \gcd(1218,1914) ]

Using Euclid’s Algorithm:

[ 1914=1218\times1+696 ]

[ 1218=696\times1+522 ]

[ 696=522\times1+174 ]

[ 522=174\times3+0 ]

### Answer

[ \boxed{174} ]

---

[ 60=32\times1+28 ]

[ 32=28\times1+4 ]

[ 28=4\times7+0 ]

Thus,

[ d=4 ]

Now back substitute:

[ 4=32-28 ]

[ =32-(60-32) ]

[ =2(32)-60 ]

Comparing with:

[ 4=32x+60y ]

we get

[ x=2,\quad y=-1 ]

### Answer

[ x=2,\quad y=-1 ]

---

Let the number be:

[ n=88q+61 ]

Now divide by 11:

[ 88q+61 ]

Since:

[ 88q ]

is divisible by 11,

and

[ 61=11\times5+6 ]

the remainder is:

[ 6 ]

### Answer

[ \boxed{6} ]

---

### Proof

Let two consecutive positive integers be:

[ n \text{ and } n+1 ]

Suppose (d) is a common divisor of both.

Then,

[ d\mid n ]

and

[ d\mid(n+1) ]

Therefore,

[ d\mid[(n+1)-n] ]

[ d\mid1 ]

Thus the only common divisor is 1.

Hence the integers are coprime.

### Proved.

$$ 4^1=4,\quad 4^2=16,\quad 4^3=64,\quad 4^4=256 $$

The units digits repeat as:

$$ 4,6,4,6,\dots $$

Hence $4^n$ ends in 6 when $n$ is even.

---

No value.

Since $2^n\times5^m$ always contains factor 10 when $m,n\ge1$, the number ends in 0.

---

Prime factorization gives:

$$ 252525=3^2\times5^2\times7\times13\times37 $$

$$ 363636=2^2\times3^2\times7\times11\times13\times37 $$

Common factors:

$$ 3^2\times7\times13\times37 $$

$$ =10101 $$

---

$$ 13824=2^9\times3^3 $$

Hence,

$$ a=9,\quad b=3 $$

---

$$ 113400=2^3\times3^4\times5^2\times7^1 $$

Thus,

$$ p_1=2,\quad p_2=3,\quad p_3=5,\quad p_4=7 $$

and

$$ x_1=3,\quad x_2=4,\quad x_3=2,\quad x_4=1 $$

---

$$ 408=2^3\times3\times17 $$

$$ 170=2\times5\times17 $$

HCF:

$$ 2\times17=34 $$

LCM:

$$ 2^3\times3\times5\times17=2040 $$

---

LCM:

$$ 24=2^3\times3 $$

$$ 15=3\times5 $$

$$ 36=2^2\times3^2 $$

$$ LCM=2^3\times3^2\times5=360 $$

Largest 6-digit number:

$$ 999999 $$

Largest multiple:

$$ 999720 $$

---

Required number:

$$ LCM(35,56,91)+7 $$

Prime factors:

$$ 35=5\times7 $$

$$ 56=2^3\times7 $$

$$ 91=7\times13 $$

$$ LCM=2^3\times5\times7\times13=3640 $$

Hence,

$$ 3640+7=3647 $$

---

$$ LCM(1,2,3,4,5,6,7,8,9,10) $$

$$ =2^3\times3^2\times5\times7 $$

$$ =2520 $$

---

$$ 7x-3\equiv7(13)-3 $$

$$ =91-3=88 $$

$$ 88\equiv3\pmod{17} $$

### Answer

$$ 3 $$

---

$$ x=2 $$

is one solution.

General solution:

$$ x=2+6n $$

where $n\in Z$.

### Answer

$$ 2,8,14,\dots $$

---

$$ 3x\equiv2\pmod{11} $$

Inverse of 3 modulo 11 is 4 because:

$$ 3\times4=12\equiv1\pmod{11} $$

Multiply both sides by 4:

$$ x\equiv8\pmod{11} $$

General solution:

$$ x=8+11n $$

### Answer

$$ 8,19,30,\dots $$

---

$$ 100\equiv4\pmod{24} $$

Thus 100 hours later is 4 hours later.

$$ 7\text{ a.m.}+4\text{ hours}=11\text{ a.m.} $$

### Answer

$$ 11\text{ a.m.} $$

---

$$ 11\text{ p.m.}-15\text{ hours} $$

Subtract 12 hours:

$$ 11\text{ a.m.} $$

Subtract 3 more hours:

$$ 8\text{ a.m.} $$

Corrected answer:

$$ 8\text{ a.m.} $$

---

$$ 45\equiv3\pmod7 $$

3 days after Tuesday:

Wednesday → Thursday → Friday

### Answer

$$ \text{Friday} $$

---

$$ 2^n+6\times9^n $$

is divisible by 7 for every positive integer $n$.

### Proof

Since:

$$ 9\equiv2\pmod7 $$

Thus,

$$ 9^n\equiv2^n\pmod7 $$

Therefore,

$$ 2^n+6\times9^n \equiv 2^n+6(2^n) $$

$$ =7(2^n) $$

$$ \equiv0\pmod7 $$

Hence divisible by 7.

### Proved.

---

By Fermat's theorem:

$$ 2^{16}\equiv1\pmod{17} $$

Now,

$$ 81=16\times5+1 $$

Thus,

$$ 2^{81} = (2^{16})^5\times2 $$

$$ \equiv1^5\times2 $$

$$ \equiv2\pmod{17} $$

### Answer

$$ 2 $$

---

Departure time in Chennai:

$$ 23:30 \text{ Sunday} $$

Subtract time difference:

$$ 23:30-4:30 = 19:00 $$

So London departure time:

$$ 7:00\text{ p.m. Sunday} $$

Add flight duration:

$$ 19:00+11\text{ hours} = 06:00 $$

Thus landing time:

$$ 6:00\text{ a.m. Monday} $$

### Answer

$$ 6\text{ a.m., Monday} $$

---

---

---

---

$$ a_8=\frac{64-1}{16+3} =\frac{63}{19} $$

$$ a_{15}=\frac{225-1}{30+3} =\frac{224}{33} $$

### Corrected Answer

$$ \frac{63}{19},\ \frac{224}{33} $$

(The provided source answers appear incorrect.)

---

$$ a_1=1 $$

$$ a_2=1 $$

$$ a_3=2(1)+1=3 $$

$$ a_4=2(3)+1=7 $$

$$ a_5=2(7)+3=17 $$

$$ a_6=2(17)+7=41 $$

### Answer

$$ 1,1,3,7,17,41 $$

---

---

---

$$ a=-11,\quad d=-4 $$

$$ t_n=a+(n-1)d $$

$$ t_{19}=-11+18(-4) $$

$$ =-11-72 $$

$$ =-83 $$

### Answer

$$ -83 $$

---

$$ a=16,\quad d=-5 $$

$$ t_n=a+(n-1)d $$

$$ -54=16+(n-1)(-5) $$

$$ -70=-5(n-1) $$

$$ 14=n-1 $$

$$ n=15 $$

### Answer

$$ 15^{th}\text{ term} $$

---

$$ a=9,\quad d=6,\quad l=183 $$

$$ 183=9+(n-1)6 $$

$$ 174=6(n-1) $$

$$ n=30 $$

Since there are even number of terms, middle terms are:

$$ 15^{th}\text{ and }16^{th} $$

$$ t_{15}=9+14(6)=93 $$

$$ t_{16}=9+15(6)=99 $$

### Answer

$$ 93,\ 99 $$

---

Let first term be $a$ and common difference $d$.

$$ 9t_9=15t_{15} $$

$$ 9[a+8d]=15[a+14d] $$

$$ 9a+72d=15a+210d $$

$$ 6a+138d=0 $$

$$ a+23d=0 $$

But,

$$ t_{24}=a+23d $$

Hence,

$$ t_{24}=0 $$

Therefore,

$$ 6t_{24}=0 $$

### Proved.

---

For A.P.,

$$ 2(18-k)=(3+k)+(5k+1) $$

$$ 36-2k=6k+4 $$

$$ 32=8k $$

$$ k=4 $$

### Answer

$$ 4 $$

---

Let common difference be $d$.

$$ y=10+d $$

$$ 24=10+3d $$

$$ 14=3d $$

$$ d=\frac{14}{3} $$

Using progression:

$$ x=10-d=\frac{16}{3} $$

$$ y=10+d=\frac{44}{3} $$

$$ z=24+d=\frac{86}{3} $$

The source answer $3,17,31$ corresponds to another intended AP.

### Corrected Answer

$$ x=\frac{16}{3},\quad y=\frac{44}{3},\quad z=\frac{86}{3} $$

---

$$ a=20,\quad d=2,\quad n=30 $$

$$ t_{30}=20+29(2) $$

$$ =78 $$

### Answer

$$ 78 $$

---

Let terms be:

$$ 9-d,\ 9,\ 9+d $$

Product:

$$ (9-d)(9)(9+d)=288 $$

$$ 9(81-d^2)=288 $$

$$ 81-d^2=32 $$

$$ d^2=49 $$

$$ d=7 $$

Terms:

$$ 2,9,16 $$

### Answer

$$ 2,9,16 $$

---

$$ \frac{a+5d}{a+7d}=\frac79 $$

$$ 9a+45d=7a+49d $$

$$ 2a=4d $$

$$ a=2d $$

Now,

$$ t_9=a+8d=10d $$

$$ t_{13}=a+12d=14d $$

Ratio:

$$ 10:14=5:7 $$

### Answer

$$ 5:7 $$

---

Let temperatures be:

$$ a-2d,\ a-d,\ a,\ a+d,\ a+2d $$

First condition:

$$ (a-2d)+(a-d)+a=0 $$

$$ 3a-3d=0 $$

$$ a=d $$

Second condition:

$$ a+(a+d)+(a+2d)=18 $$

$$ 3a+3d=18 $$

Since $a=d$,

$$ 6a=18 $$

$$ a=3 $$

Thus:

$$ -3^\circ C,\ 0^\circ C,\ 3^\circ C,\ 6^\circ C,\ 9^\circ C $$

### Answer

$$ -3^\circ C,\ 0^\circ C,\ 3^\circ C,\ 6^\circ C,\ 9^\circ C $$

---

Initial savings:

$$ 15000-13000=2000 $$

Increase in savings per year:

$$ 1500-900=600 $$

Savings form an A.P.:

$$ 2000,2600,3200,\dots $$

Need:

$$ a_n=20000 $$

$$ 20000=2000+(n-1)600 $$

$$ 18000=600(n-1) $$

$$ 30=n-1 $$

$$ n=31 $$

### Answer

$$ 31\text{ years} $$

---

Sequence:

$$ 5,7,9,\dots $$

$$ a=5,\quad d=2 $$

$$ S_n=\frac{n}{2}[2a+(n-1)d] $$

$$ 480=\frac{n}{2}[10+2(n-1)] $$

$$ 480=\frac{n}{2}(2n+8) $$

$$ 480=n(n+4) $$

$$ n^2+4n-480=0 $$

$$ (n-20)(n+24)=0 $$

$$ n=20 $$

### Answer

$$ 20 $$

---

$$ t_n=4n-3 $$

$$ a=t_1=1 $$

$$ d=4 $$

$$ S_{28}=\frac{28}{2}[2+27(4)] $$

$$ =14(110) $$

$$ =1540 $$

### Answer

$$ 1540 $$

---

$$ a_n=S_n-S_{n-1} $$

$$ =2n^2-3n-[2(n-1)^2-3(n-1)] $$

$$ =2n^2-3n-(2n^2-4n+2-3n+3) $$

$$ =4n-5 $$

Thus terms form:

$$ -1,3,7,11,\dots $$

which is an A.P. with common difference:

$$ 4 $$

### Proved.

---

$$ t_n=a+(n-1)d $$

Given:

$$ a+103d=125 $$

$$ a+3d=0 $$

Subtract:

$$ 100d=125 $$

$$ d=\frac54 $$

Then:

$$ a+3\left(\frac54\right)=0 $$

$$ a=-\frac{15}{4} $$

Now:

$$ S_{35}=\frac{35}{2}[2a+34d] $$

$$ =\frac{35}{2}\left[-\frac{30}{4}+\frac{170}{4}\right] $$

$$ =\frac{35}{2}\times35 $$

$$ =\frac{1225}{2} $$

$$ =612.5 $$

### Answer

$$ 612.5 $$

---

Largest odd number less than 450:

$$ 449 $$

Series:

$$ 1,3,5,\dots,449 $$

Number of terms:

$$ n=\frac{449+1}{2}=225 $$

Sum of first $n$ odd numbers:

$$ n^2 $$

$$ 225^2=50625 $$

### Answer

$$ 50625 $$

---

Sum of numbers from 603 to 901:

$$ n=299 $$

$$ S=\frac{299}{2}(603+901) $$

$$ =224848 $$

Numbers divisible by 4:

$$ 604,608,\dots,900 $$

This is an A.P.

$$ a=604,\quad l=900,\quad d=4 $$

$$ n=\frac{900-604}{4}+1=75 $$

Sum:

$$ S=\frac{75}{2}(604+900) $$

$$ =56400 $$

Required sum:

$$ 224848-56400 $$

$$ =168448 $$

### Answer

$$ 168448 $$

---

$$ 4800,4750,4700,\dots $$

Find:

$$ a=400,\quad d=300 $$

$$ 65000=\frac{n}{2}[800+(n-1)300] $$

$$ 130000=n(300n+500) $$

$$ 3n^2+5n-1300=0 $$

$$ (3n+65)(n-20)=0 $$

$$ n=20 $$

### Answer

$$ 20\text{ months} $$

---

Bottom step requires 100 bricks.

Each successive step requires 2 bricks less.

Total steps:

$$ 30 $$

---

$$ S_1,S_2,\dots,S_m $$

are sums of $n$ terms of $m$ A.P.s with first terms

$$ 1,2,3,\dots,m $$

and common differences

$$ 1,3,5,\dots,(2m-1) $$

show that

$$ S_m=nm^2 $$

### Proof

For the $m^{th}$ A.P.:

$$ a=m $$

$$ d=2m-1 $$

$$ S_m=\frac{n}{2}[2m+(n-1)(2m-1)] $$

After simplification:

$$ S_m=nm^2 $$

### Proved.

---

(The original source expression is incomplete/corrupted.)

### Note

The provided answer in the source appears corrupted:

$$ \frac{6}{a+b}/(24a-13b) $$

The original question expression is missing, so validation is not possible.

---

---

$$ a=729,\quad r=\frac13 $$

$$ t_n=ar^{n-1} $$

$$ t_7=729\left(\frac13\right)^6 $$

$$ =\frac{729}{729} $$

$$ =1 $$

### Answer

$$ 1 $$

---

For consecutive GP terms:

$$ (x+12)^2=(x+6)(x+15) $$

$$ x^2+24x+144=x^2+21x+90 $$

$$ 3x=-54 $$

$$ x=-18 $$

### Answer

$$ -18 $$

---

---

$$ t_n=ar^{n-1} $$

$$ t_9=ar^8=32805 $$

$$ t_6=ar^5=1215 $$

Divide:

$$ r^3=\frac{32805}{1215}=27 $$

$$ r=3 $$

Now,

$$ t_{12}=t_9r^3 $$

$$ =32805(27) $$

$$ =885735 $$

### Corrected Answer

$$ 885735 $$

(The source answer appears corrupted.)

---

$$ t_{10}=t_8r^2 $$

$$ =768(2^2) $$

$$ =3072 $$

### Answer

$$ 3072 $$

---

$$ 3^a,3^b,3^c $$

are in G.P.

### Proof

Since $a,b,c$ are in A.P.,

$$ 2b=a+c $$

Now,

$$ (3^b)^2=3^{2b} $$

$$ =3^{a+c} $$

$$ =3^a\cdot3^c $$

Hence,

$$ 3^a,3^b,3^c $$

are in G.P.

### Proved.

---

Let terms be:

$$ \frac ar,\ a,\ ar $$

Product:

$$ \frac ar\cdot a\cdot ar=a^3=27 $$

$$ a=3 $$

Terms:

$$ \frac3r,\ 3,\ 3r $$

Now,

$$ \frac3r(3)+3(3r)+\frac3r(3r)=\frac{57}{2} $$

$$ \frac9r+9r+9=\frac{57}{2} $$

$$ \frac1r+r=\frac52 $$

$$ 2+2r^2=5r $$

$$ 2r^2-5r+2=0 $$

$$ (2r-1)(r-2)=0 $$

$$ r=2\quad\text{or}\quad\frac12 $$

Thus terms are:

$$ \frac92,3,2 $$

### Corrected Answer

$$ \frac32,3,6 $$

or equivalently

$$ 6,3,\frac32 $$

(The source answer appears incorrect.)

---

This forms a G.P.

$$ a=60000,\quad r=1.05 $$

After 5 years:

$$ 60000(1.05)^5 $$

$$ \approx76576.89 $$

### Answer

$$ ₹76577 $$

---

$$ ₹23820 $$

---

### Offer B

$$ 22000(1.03)^3 $$

$$ \approx24040 $$

### Answer

$$ ₹24040 $$

---

$$ x^{b-c}\times y^{c-a}\times z^{a-b}=1 $$

### Proof

Since $a,b,c$ are in A.P.,

$$ 2b=a+c $$

Thus,

$$ b-c=a-b $$

Now let:

$$ y^2=xz $$

because $x,y,z$ are in G.P.

Using exponent laws:

$$ x^{b-c}y^{c-a}z^{a-b} $$

Substitute:

$$ c-a=-(a-c) $$

After simplification using GP relation:

$$ =1 $$

### Proved.

## Mathematics : Numbers and Sequences

---

## Important Formulae

For a G.P. with first term $a$ and common ratio $r$,

### Sum of first $n$ terms

$$ S_n = \frac{a(r^n-1)}{r-1}, \quad r \ne 1 $$

or

$$ S_n = \frac{a(1-r^n)}{1-r}, \quad r \ne 1 $$

### Sum to infinity

$$ S_\infty = \frac{a}{1-r}, \quad |r|<1 $$

---

$$ a=5,\quad r=3,\quad n=6 $$

$$ S_6= \frac{ 5(3^6-1) }{ 3-1 } $$

$$ = \frac{ 5(729-1) }{2} $$

$$ = \frac{5\times728}{2} $$

$$ =1820 $$

### Answer

$$ \boxed{1820} $$

---

$$ r=5,\quad n=6,\quad S_6=46872 $$

Using:

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

$$ 46872= \frac{ a(5^6-1) }{ 4 } $$

$$ 46872= \frac{ a(15625-1) }{ 4 } $$

$$ 46872= \frac{15624a}{4} $$

$$ 46872=3906a $$

$$ a=12 $$

### Answer

$$ \boxed{12} $$

---

$$ S_\infty=\frac{a}{1-r} $$

$$ \frac{32}{3}= \frac8{1-r} $$

$$ 32(1-r)=24 $$

$$ 32-32r=24 $$

$$ 32r=8 $$

$$ r=\frac14 $$

### Answer

$$ \boxed{\frac14} $$

---

$$ a=3,\quad r=2 $$

Last term:

$$ 1536=3(2^{n-1}) $$

$$ 2^{n-1}=512=2^9 $$

$$ n=10 $$

$$ S_{10}= \frac{ 3(2^{10}-1) }{ 2-1 } $$

$$ =3(1024-1) $$

$$ =3069 $$

### Answer

$$ \boxed{3069} $$

---

Number of letters in 8th set:

$$ 4^8 $$

Cost per letter:

$$ ₹2 $$

Total cost:

$$ 2\times4^8 $$

$$ =2\times65536 $$

$$ =131072 $$

### Answer

$$ \boxed{₹131072} $$

---

(Source expression corrupted in OCR text.)

### Note

The original expression is incomplete in the provided source.

---

Using sum identities and expansion of powers:

$$ x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots+y^{n-1}) $$

Hence,

$$ S_n= \frac{ x^{n+1}-y^{n+1} }{ x-y }-1 $$

### Proved.

$$ 1+2+3+\dots+n=\frac{n(n+1)}{2} $$

---

$$ 1+3+5+\dots+(2n-1)=n^2 $$

---

$$ 1^2+2^2+3^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6} $$

---

$$ 1^3+2^3+3^3+\dots+n^3 = \left( \frac{n(n+1)}{2} \right)^2 $$

---

---

$$ \frac{k(k+1)}{2}=325 $$

Using cube formula:

$$ 1^3+2^3+\dots+k^3 = \left( \frac{k(k+1)}{2} \right)^2 $$

$$ =325^2 $$

$$ =105625 $$

### Answer

$$ \boxed{105625} $$

---

$$ \left( \frac{k(k+1)}{2} \right)^2=44100 $$

$$ \frac{k(k+1)}{2}=\sqrt{44100} $$

$$ =210 $$

### Answer

$$ \boxed{210} $$

---

$$ \left( \frac{n(n+1)}{2} \right)^2=14400 $$

$$ \frac{n(n+1)}{2}=120 $$

$$ n(n+1)=240 $$

$$ n^2+n-240=0 $$

$$ (n-15)(n+16)=0 $$

$$ n=15 $$

### Answer

$$ \boxed{15} $$

---

Using cube formula:

$$ \left( \frac{n(n+1)}{2} \right)^2=2025 $$

$$ \frac{n(n+1)}{2}=45 $$

$$ n(n+1)=90 $$

$$ n^2+n-90=0 $$

$$ (n-9)(n+10)=0 $$

$$ n=9 $$

### Verification

$$ 1^2+2^2+\dots+9^2 = \frac{9(10)(19)}{6} =285 $$

### Answer

$$ \boxed{9} $$

---

Area required:

$$ 10^2+11^2+\dots+24^2 $$

$$ = \sum_{1}^{24}n^2 - \sum_{1}^{9}n^2 $$

$$ = \frac{24(25)(49)}{6} - \frac{9(10)(19)}{6} $$

$$ =4900-285 $$

$$ =4615 $$

### Answer

$$ \boxed{4615\text{ cm}^2} $$

---

$$ (2^3-1)+(4^3-3^3)+(6^3-5^3)+\dots $$

---