32nd term

 No

\(\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}\)
∆ABC∼ΔDEF
⇒ \(\frac{area\Delta ABC}{area\Delta DEF}=\frac{AB^{2}}{DE^{2}}\)
=\(\left(\frac{AB}{DE}\right)^{2}=\left(\frac{4}{7}\right)^{2}=\frac{16}{49}\)

In \(\triangle\)ABC from Pythagoras theorem,

\(\Rightarrow\) AB² = AD² + BD²
\(\Rightarrow\) BC² = AD² + BD² (as AB = BC = CA)
\(\Rightarrow\) (2BD)² = AD² + BD², ( \(\bot \)is the median in an equilateral)
\(\therefore\) 3BD² = AD².

Let \(CD\bot AB,\)
then CD=p
Area of \(\triangle\)ABC= \(\frac {1}{2}\) x base x height
\(\Rightarrow\)Area of \(\triangle\)ABC=\(\frac {1}{2}\)  x AB x CD= cp
Also, Area of \(\triangle\)ABC=\(\frac {1}{2}\) x BC x AC= ab
\(\frac {1}{2}\)cp=\(\frac {1}{2}\)ab
\(\Rightarrow\)cp=ab.

Since, XY || QR
\(\therefore \frac { PX }{ XQ } =\frac { PY }{ YR } \)
\(\Rightarrow \frac { 1 }{ 2 } =\frac { PY }{ PR-PY } \)
\(\Rightarrow\) PR-4=8
\(\Rightarrow\) PR=12 cm

In right \(\triangle\)PQR, QR²=PR²-PQ²
= 12²-6²
= 144-36=108
QR = 6\(\sqrt3\)cm

In \(\triangle\)ABC and \(\triangle\)DEF,
AB = 3DE and AB = 3DF
\(\Rightarrow \frac { AB }{ DE } =3; \frac { AC }{ DF } =3;\)
P₁ = 3P₂
BC = 3EF
\(\Rightarrow \frac { AB }{ DE } =\frac { AC }{ DF } =\frac { BC }{ EF } =3\)
\(\triangle\)ABC~\(\triangle\)DEF
\(\Rightarrow\frac { ar(\triangle ABC) }{ ar(\triangle DEF) } ={ \left( \frac { AB }{ DE } \right) }^{ 2 }={ (3) }^{ 2 }=9.\)

PQ || MN
So, \(\frac { KP }{ PM } =\frac { KQ }{ QN } \)
\(\Rightarrow \quad \frac { KP }{ PM } =\frac { KQ }{ KN-KQ } \)
\(\Rightarrow \quad \frac { 4 }{ 13 } =\frac { KQ }{ 20.4-KQ } \)
\(\Rightarrow \) 4 x 20.4-4KQ=13 KQ
\(\Rightarrow \)17 KQ=4 x 20.4
\(\therefore \quad KQ=\frac { 20.4\times 4 }{ 17 } =4.8\quad cm,\)

No, Angle included should be same.

Hence HCF (1,656,4,025)= 23

\(a=\frac { 9,009 }{ 3,003 } =3\)
\(b=\frac { 1,001 }{ 143 } =7\)
Since 143 = 11 x 13, so c=11 or 13
and  d=13 or 11.

HCF of K
k.2
k.3
k.2²
k.5 is k

The smallest prime number is 2 and the smallest composite number is 2² Hence, required HCF (2²,2) = 2.

The given number ends in 5. Hence it is a multiple of 5. Therefore it is a composite number

\(\frac{16}{9}\)

Let h be the height of the tower.

[\(\because\)  80 cm = 0.8 m]
Since, it is clear that both the triangles will be similar.
\(\therefore \frac{1}{0.8}=\frac{h}{30} \quad \Rightarrow h = \frac{30}{0.8}\) = 37.5 m

3¹²

No, because 6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
If the number 6ⁿ ends with digit 5, then it will be divisible by 5, i.e. its one factor will be 5. But the only primes in the factorisation of 6ⁿ are 2 and 3, but not 5.
Hence, it cannot end with digit 5.

Let y = \(0.5\overline { 4 } \) = 0.54444...
On multiplying both sides by 10, we get
10 y = 5.4444  ....(i)
On multiplying Eq.(i)  by 10, we get
100 y = 54.4444    ....(ii)
On subtracting Eq.(i) from Eq.(ii), we get
90 y = 49 \(\Rightarrow \quad y=\frac { 49 }{ 90 } \)

Here, a = 3 and d = 10 - 3 = 7
a₁₃ = a + 12d
= 3 + 12 X 7 = 87
Let a_n is84 more than 13th terms.
a_n = a₁₃ + 84
\(\Rightarrow\)  a + ( n - 1 )d = 87 + 84
\(\Rightarrow\)  3 + ( n - 1 )7 = 171
\(\Rightarrow\)    3 + 7n - 7 = 171
\(\Rightarrow\)    7n = 175
\(\Rightarrow\)      n = 25
\(\therefore\)   25th term is 84 more than 13th term.

Let the 1st term of the AP = a and common difference = d
Then \(S_{ 6 }\frac { 6 }{ 2 } \left\{ 2a+(6-1d) \right\} \) 
\(\Rightarrow 42=3(2a+5d)\)
\(\Rightarrow 2a+5d=14\)   ...(i)
Also, \(\frac { a_{ 10 } }{ a_{ 30 } } =\frac { 1 }{ 3 } \)
\(\Rightarrow \frac { a+9d }{ a+29d } =\frac { 1 }{ 3 } \)
\(\Rightarrow \) 3a+27d=a+29d
\(\Rightarrow \) 2a=2d \(\Rightarrow \) a=d
Substituting the value of d in equation (i)
we get
2a+5a=14 \(\Rightarrow \) a=2
\(\therefore \) d=2
Now a₁₃ = a+12d
\(\Rightarrow \) a₁₃ =2+12x2=26
\(\therefore \) 1st term =2
Thirteenth term =26

Two-digit positive numbers divisible by
3 are 12, 15, 18, ..., 99
These numbers are in A.P. with a - 12,
d = 15 - 12 = 3 and a_n = 99
\(\therefore\)   a + ( n - 1 )d = 99
\(\Rightarrow\)   12 + ( n - 1 )3 = 99
\(\Rightarrow\)   ( n - 1 )3 = 87
\(\Rightarrow\)     n ​​​​​​​- 1 = 29  \(\Rightarrow\)  n = 30
Now   S_n = \(\frac{n}{2}(a+l)\)
\(\Rightarrow\)    S₃₀ = \(\frac{30}{2}(12+99)=1665 \)

 S_n = \(\frac{5{n}^{2}}{2}+\frac{3n}{2}\)
\(\Rightarrow\)     S1 = \(\frac{5\times{1}^{2}}{2}+\frac{3\times1}{2}=4={a}_{1}\)
S2 = \(\frac{5\times{2}^{2}}{2}+\frac{3\times2}{2}=13\)
\(\Rightarrow\)   a₁ + a₂ = 13  \(\Rightarrow\) 4 + a₂ = 13         [Using eq.(i)]
\(\Rightarrow\)  a₂ = 9
d = a₂ - a₁ = 9 - 4 = 5
Now,   a₂₀ = a + 19d = 4 + 19 x 5 = 99.

S_n = 4n - n²
Puttuing n = 1, 2, 3, ...
S₁ = 4 X 1 - 1² = 4 - 1 = 3
a₁ = 3
S₂ = 4(2) - 2² = 8 - 4 = 4
a₁ + a₂ = 4
\(\Rightarrow\)     3 + a₂ = 4
\(\Rightarrow\)    a₂ = 4 - 3 = 1
d = a₂ - a₁ = 1 - 3 = - 2
a₃ = a₂ + d = 1 (-2) = -1
a₁₀ = a + 9d = 3 + 9 (-2)= 3 - 18 = - 15
a_n = a + ( n - 1 ) d = 3 + ( n - 1 ) ( -2 ) = 3 - 2n + 2 = 5 - 2n

Let Is term = a, common difference = d.
a₁₀ = a + 9d, a₂₄ = a + 23d
According to the question, a₂₄ = 2 X a₁₀
\(\Rightarrow\) a + 23d = 2 ( a + 9d ) \(\Rightarrow\) a + 23d = 2a + 18d \(\Rightarrow\) a = 5d
Now,  a₁₆ = a + 15d = 5d + 15d = 20d     ..(i)
a₃₆ = a + 35d = 5d + 35d = 40d         ..(ii)
From (i) and (ii), we get
a₃₆ = 2 x a₁₆                 
 Hence proved.

Here, a = 21,  d = 18 - 21 = -3
Let    a_n= 0
\(\Rightarrow\)   a + (n - 1)d = 0
\(\Rightarrow\)   21 + (n - 1)(-3) = 0
\(\Rightarrow\) (n - 1)(-3) = -21
\(\Rightarrow\) n - 1 =\(\frac{-21}{-3}\)= 7
\(\Rightarrow\) n = 8
8th term is zero.

Given;     \({ a }_{ 11 }=38\) and \({ a }_{ 16 }=73\)
 \(a + 10d = 38\) and \(a + 15d = 73\)
\(\Rightarrow\)  \(a + 15d - a - 10d = 73 - 38\)
 \(5d = 35\)
 \(d =\)\(\frac{35}{5}\) \(= 7\)
\({ a }_{ 11 }=a+10x7=38\) \(\Rightarrow\) \(a = -32\)
\({ a }_{ 31 }=a+30d=178\)