Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 5 - Arithmetic Progression

Answer:

(i) The given progression 9, 15, 21, 27, ... .

Clearly, 15 − 9 = 21 − 15 = 27 − 21 = 6 (Constant)

Thus, each term differs from its preceding term by 6. So, the given progression is an AP.

First term = 9

Common difference = 6

Next term of the AP = 27 + 6 = 33

(ii) The given progression 11, 6, 1, −4, ... .

Clearly, 6 − 11 = 1 − 6 = −4 − 1 = −5 (Constant)

Thus, each term differs from its preceding term by −5. So, the given progression is an AP.

First term = 11

Common difference = −5

Next term of the AP = −4 + (−5) = −9

(iii) The given progression −1, $\frac{-5}{6}$, $\frac{-2}{3}$, $\frac{-1}{2}$, ...

Clearly, $\frac{-5}{6}-\left(-1\right)=\frac{-2}{3}-\left(\frac{-5}{6}\right)=\frac{-1}{2}-\left(\frac{-2}{3}\right)=\frac{1}{6}$ (Constant)

Thus, each term differs from its preceding term by $\frac{1}{6}$. So, the given progression is an AP.

First term = −1

Common difference = $\frac{1}{6}$

Next term of the AP = $\frac{-1}{2}+\frac{1}{6}=\frac{-2}{6}=\frac{-1}{3}$

(iv) The given progression $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, ...$

This sequence can be re-written as $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, ...$

Clearly, $2\sqrt{2}-\sqrt{2}=3\sqrt{2}-2\sqrt{2}=4\sqrt{2}-3\sqrt{2}=\sqrt{2}$ (Constant)

Thus, each term differs from its preceding term by $\sqrt{2}$. So, the given progression is an AP.

First term = $\sqrt{2}$

Common difference = $\sqrt{2}$

Next term of the AP = $4\sqrt{2}+\sqrt{2}=5\sqrt{2}=\sqrt{50}$

(v) The given progression $\sqrt{20}, \sqrt{45}, \sqrt{80}, \sqrt{125}, ...$

This sequence can be re-written as $2\sqrt{5}, 3\sqrt{5}, 4\sqrt{5}, 5\sqrt{5}, ...$

Clearly, $3\sqrt{5}-2\sqrt{5}=4\sqrt{5}-3\sqrt{5}=5\sqrt{5}-4\sqrt{5}=\sqrt{5}$ (Constant)

Thus, each term differs from its preceding term by $\sqrt{5}$. So, the given progression is an AP.

First term = $2\sqrt{5}=\sqrt{20}$

Common difference = $\sqrt{5}$

Next term of the AP = $5\sqrt{5}+\sqrt{5}=6\sqrt{5}=\sqrt{180}$

Answer:

(i)  The given AP is 9, 13, 17, 21, ... .

First term, a = 9

Common difference, d = 13 − 9 = 4

n^th term of the AP, a_n = a + (n − 1)d = 9 + (n − 1) × 4

∴ 20th term of the AP, a₂₀ = 9 + (20 − 1) × 4 = 9 + 76 = 85

(ii)  The given AP is 20, 17, 14, 11, ... .

First term, a = 20

Common difference, d = 17 − 20 = −3

n^th term of the AP, a_n = a + (n − 1)d = 20 + (n − 1) × (−3)

∴ 35th term of the AP, a₃₅ = 20 + (35 − 1) × (−3) = 20 − 102 = −82

(iii)  The given AP is $\sqrt{2}, \sqrt{18}, \sqrt{50}, \sqrt{98}$, ... .

This can be re-written as $\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}, 7\sqrt{2}$, ... .

First term, a = $\sqrt{2}$

Common difference, d = $3\sqrt{2}-\sqrt{2}=2\sqrt{2}$

n^th term of the AP, a_n = a + (n − 1)d = $\sqrt{2}+\left(n-1\right)\times 2\sqrt{2}$

∴ 18th term of the AP, a₁₈ = $\sqrt{2}+\left(18-1\right)\times 2\sqrt{2}=\sqrt{2}+34\sqrt{2}=35\sqrt{2}=\sqrt{2450}$

(iv)  The given AP is $\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4}$, ... .

First term, a = $\frac{3}{4}$

Common difference, d = $\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}$

n^th term of the AP, a_n = a + (n − 1)d = $\frac{3}{4}+\left(n-1\right)\times \left(\frac{1}{2}\right)$

∴ 9th term of the AP, a₉ = $\frac{3}{4}+\left(9-1\right)\times \frac{1}{2}=\frac{3}{4}+4=\frac{19}{4}$

(v)  The given AP is −40, −15, 10, 35, ... .

First term, a = −40

Common difference, d = −15 − (−40) = 25

n^th term of the AP, a_n = a + (n − 1)d = −40 + (n − 1) × 25

∴ 15th term of the AP, a₁₅ = −40 + (15 − 1) × 25 = −40 + 350 = 310

Answer:

(i) The given AP is $6, 7\frac{3}{4}, 9\frac{1}{2}, 11\frac{1}{4},...$ 
First term, a = 6 and common difference, d $= 7\frac{3}{4} - 6 \Rightarrow \frac{31}{4} -6 \Rightarrow \frac{31 - 24}{4} = \frac{7}{4}$
$$\begin{gathered} \mathrm{Now}, T_{37} = a+(37-1) d = a+36d \\ =6+36\times \frac{7}{4}=6+63=69 \\ \therefore {37}^{th} \mathrm{term}=69 \end{gathered}$$

(ii) The given AP is $5, 4\frac{1}{2}, 4, 3\frac{1}{2}, 3,...$ 
    First term = 5
    Common difference$= 4\frac{1}{2} - 5 \Rightarrow \frac{9}{2} -5 \Rightarrow \frac{9 - 10}{2} = -\frac{1}{2}$
    $$\begin{gathered} \therefore a =5 \mathrm{and} d=-\frac{1}{2} \\ \mathrm{Now}, T_{25 }= a+(25-1)d=a+24d \\ =5+24\times \left(-\frac{1}{2}\right)=5-12=-7 \\ \therefore {25}^{th} \mathrm{term}=-7 \end{gathered}$$

Answer:

$$\begin{gathered} 2p-1,3p+1,11 \mathrm{are} \mathrm{in} \mathrm{AP} \\ \mathrm{A}.\mathrm{P} \mathrm{has} \mathrm{common} \mathrm{difference} \\ d=a_2-a_1 \\ \mathrm{Here}, a_1=2p-1,a_2=3p+1 \mathrm{and} a_3=11 \\ {a}_2-a_1=a_3-a_2 \\ 3p+1-\left(\mathit{2}p-\mathit{1}\right)=11-3p-1 \\ p+2=10-3p \\ \Rightarrow p=2 \end{gathered}$$
And the numbers are:
$$\begin{gathered} 2\left(2\right)-1, 3\left(2\right)+1, 11 \\ \Rightarrow 3, 7, 11 \mathrm{are} \mathrm{the} \mathrm{numbers} \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ 5,11,17,23,... \\ {a}_n=a+\left(n-1\right)d \\ a=5 \\ d=11-5=6 \\ {a}_n=5+\left(n-1\right)6 \\ {a}_n=6n-1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \\ 16,9,2,-5 \\ {a}_n=a+\left(n-1\right)d \\ a=16 \\ d=9-16=-7 \\ {a}_n=16+\left(n-1\right)\left(-7\right) \\ {a}_n=16-7n+7 \\ {a}_n=23-7n \end{gathered}$$

Answer:

Tn = (4n - 10)     [Given]
​T₁ = (4 ⨯ 1 - 10) = -6
T₂ = (4 ⨯ 2 - 10) = -2
T₃ = (4 ⨯ 3 - 10) = 2
T₄ = (4 ⨯ 4 - 10) = 6

Answer:

In the given AP, a  = 6 and d = (10 - 6) = 4
Suppose that there are n terms in the given AP.
Then T_n = 174
⇒ a + (n - 1)d = 174
⇒ 6 + (n - 1) ⨯ 4 = 174
⇒ 2 + 4n = 174
⇒ 4n = 172
⇒ n = 43
Hence, there are 43 terms in the given AP.
             

Answer:

In the given AP, a  = 41 and d = (38 - 41) = -3
Suppose that there are n terms in the given AP.
Then T_n = 8
⇒ a + (n - 1) d = 8
⇒ 41 + (n - 1) ⨯ (-3) = 8
⇒ 44 - 3n = 8
⇒ 3n = 36
⇒ n = 12
Hence, there are 12 terms in the given AP.

Answer:

The given AP is 18, $15\frac{1}{2}$, 13, ..., −47.

First term, a = 18

Common difference, d = $15\frac{1}{2}-18=\frac{31}{2}-18=\frac{31-36}{2}=-\frac{5}{2}$

Suppose there are n terms in the given AP. Then,

$$\begin{gathered} a_n=-47 \\ \Rightarrow 18+\left(n-1\right)\times \left(-\frac{5}{2}\right)=-47 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow -\frac{5}{2}\left(n-1\right)=-47-18=-65 \\ \Rightarrow n-1=-65\times \left(-\frac{2}{5}\right)=26 \end{gathered}$$
$\Rightarrow n=26+1=27$

Hence, there are 27 terms in the given AP.

Answer:

In the given AP, first term, a = 3 and common difference, d = (8 - 3) = 5.

Let's its n^th term be 88.
Then T_n = 88
⇒ a + (n - 1)d = 88
⇒ 3 + (n - 1) ⨯ 5 = 88
⇒ 5n - 2 = 88
⇒ 5n = 90
⇒ n = 18
Hence, the 18^th term of the given AP is 88.

Answer:

In the given AP, first term, a = 72 and common difference, d = (68 - 72) = -4.
Let its n^th term be 0.
Then, T_n = 0  
⇒ a + (n - 1)d = 0
⇒ 72 + (n - 1) ⨯ (-4) = 0
⇒ 76 - 4n = 0
⇒ 4n = 76
⇒ n = 19
Hence, the 19^th term of the given AP is 0.

Answer:

In the given AP, first term = $\frac{5}{6}$ and common difference, d $=\left(1- \frac{5}{6} = \frac{1}{6}\right)$.
Let its n^th term be 3.

$$\begin{gathered} \mathrm{Now}, T_n=3 \\ \Rightarrow a + (n-1)d = 3 \\ \Rightarrow \frac{5}{6} + (n-1)\times \frac{1}{6} = 3 \\ \Rightarrow \frac{2}{3} + \frac{n}{6} = 3 \\ \Rightarrow \frac{n}{6} = \frac{7}{3} \\ \Rightarrow n = 14 \end{gathered}$$

Hence, the 14^th term of the given AP is 3.

Answer:

The given AP is 21, 18, 15, ... .

First term, a = 21

Common difference, d = 18 − 21 = −3

Suppose n^th term of the given AP is −81. Then,

$$\begin{gathered} a_n=-81 \\ \Rightarrow 21+\left(n-1\right)\times \left(-3\right)=-81 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow -3\left(n-1\right)=-81-21=-102 \\ \Rightarrow n-1=\frac{102}{3}=34 \end{gathered}$$
$\Rightarrow n=34+1=35$

Hence, the 35th term of the given AP is −81.

Answer:

In the given problem, let us first find the 41^st term of the given A.P.

A.P. is 8, 14, 20, 26 …

Here,

First term (a) = 8

Common difference of the A.P. (d) = 6

Now, as we know,

So, for 41^st term (n = 41),

Let us take the term which is 72 more than the 41^st term as a_n. So,

Also, 

Further simplifying, we get,

Therefore, the of the given A.P. is 72 more than the 41^st term.

Answer:

Here, a = 5 and d = (15 - 5) = 10
The 31^st term is given by
T₃₁ = a + (31 - 1)d = a + 30d = 5 + 30 ⨯ 10 = 305
∴ Required term = (305 + 130) = 435
Let this be be the n^th term.
Then T_n = 435
​⇒ 5 + (n - 1) ⨯ 10 = 435
⇒ 10n = 440
⇒ n = 44
Hence, the 44^th term will be 130 more than its 31^st term.  

Answer:

In the given AP, let the first term be a and the common difference be d.
Then, T_n = a + (n - 1)d ​
Now, we have:
T₁₀ = a + (10 - 1)d 
⇒ a + 9d  = 52       ...(1)
T₁₃ = a + (13 - 1)d = a + 12d               ...(2)
T₁₇ = a + (17 - 1)d = a + 16d                 ...(3)     

But, it is given that T₁₇ = 20 + T₁₃
i.e., a + 16d  = 20 + a + 12d 
⇒ 4d = 20  
⇒ d = 5
On substituting d = 5 in (1), we get:
a + 9 ⨯ 5 = 52 
⇒​ a = 7

Answer:

The given AP is 6, 13, 20, ..., 216. 

First term, a = 6

Common difference, d = 13 − 6 = 7

Suppose there are n terms in the given AP. Then,

$$\begin{gathered} a_n=216 \\ \Rightarrow 6+\left(n-1\right)\times 7=216 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 7\left(n-1\right)=216-6=210 \\ \Rightarrow n-1=\frac{210}{7}=30 \end{gathered}$$
$\Rightarrow n=30+1=31$

Thus, the given AP contains 31 terms.

∴ Middle term of the given AP

= $\left(\frac{31+1}{2}\right)$th term

= 16th term

= 6 + (16 − 1) × 7

= 6 + 105

= 111
 
Hence, the middle term of the given AP is 111.

Answer:

The given AP is 10, 7, 4, ..., −62.

First term, a = 10

Common difference, d = 7 − 10 = −3

Suppose there are n terms in the given AP. Then,

$$\begin{gathered} a_n=-62 \\ \Rightarrow 10+\left(n-1\right)\times \left(-3\right)=-62 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow -3\left(n-1\right)=-62-10=-72 \\ \Rightarrow n-1=\frac{72}{3}=24 \end{gathered}$$
$\Rightarrow n=24+1=25$

Thus, the given AP contains 25 terms.

∴ Middle term of the given AP

= $\left(\frac{25+1}{2}\right)$th term

= 13th term

= 10 + (13 − 1) × (−3)

= 10 − 36

= −26
 
Hence, the middle term of the given AP is −26.

Answer:

The given AP is $-\frac{4}{3},-1,\frac{-2}{3},...,4\frac{1}{3}$.

First term, a = $-\frac{4}{3}$

Common difference, d = $-1-\left(-\frac{4}{3}\right)=-1+\frac{4}{3}=\frac{1}{3}$

Suppose there are n terms in the given AP. Then,

$$\begin{gathered} a_n=4\frac{1}{3} \\ \Rightarrow -\frac{4}{3}+\left(n-1\right)\times \left(\frac{1}{3}\right)=\frac{13}{3} \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow \frac{1}{3}\left(n-1\right)=\frac{13}{3}+\frac{4}{3}=\frac{17}{3} \\ \Rightarrow n-1=17 \end{gathered}$$
$\Rightarrow n=17+1=18$

Thus, the given AP contains 18 terms. So, there are two middle terms in the given AP.

The middle terms of the given AP are $\left(\frac{18}{2}\right)$th term and $\left(\frac{18}{2}+1\right)$th term i.e. 9th term and 10th term.

∴ Sum of the middle most terms of the given AP

= 9th term + 10th term

$$\begin{gathered} =\left[-\frac{4}{3}+\left(9-1\right)\times \frac{1}{3}\right]+\left[-\frac{4}{3}+\left(10-1\right)\times \frac{1}{3}\right] \\ =-\frac{4}{3}+\frac{8}{3}-\frac{4}{3}+3 \\ =3 \end{gathered}$$
 
Hence, the sum of the middle most terms of the given AP is 3.

Answer:

Here, a = 7 and d = (10 - 7) = 3, l = 184 and n = 8^th from the end.
Now, n^th term from the end = [l - (n -1)d]
8^th term from the end = [184 - (8 - 1) ⨯ 3]
                                 = [184 - (7 ⨯ 3)] = (184 - 21) = 163
Hence, the 8^th term from the end is 163.

Answer:

Answer:

The given AP is 3, 7, 11, 15, ... .

Here, a = 3 and d = 7 − 3 = 4

Let the n^th term of the given AP be 184. Then,

$$\begin{gathered} a_n=184 \\ \Rightarrow 3+\left(n-1\right)\times 4=184 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 4n-1=184 \\ \Rightarrow 4n=185 \end{gathered}$$
$\Rightarrow n=\frac{185}{4}=46\frac{1}{4}$

But, the number of terms cannot be a fraction.

Hence, 184 is not a term of the given AP.

Answer:

The given AP is 11, 8, 5, 2, ... .

Here, a = 11 and d = 8 − 11 = −3

Let the n^th term of the given AP be −150. Then,

$$\begin{gathered} a_n=-150 \\ \Rightarrow 11+\left(n-1\right)\times \left(-3\right)=-150 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow -3n+14=-150 \\ \Rightarrow -3n=-164 \end{gathered}$$
$\Rightarrow n=\frac{164}{3}=54\frac{2}{3}$

But, the number of terms cannot be a fraction.

Hence, −150 is not a term of the given AP.

Answer:

The given AP is 121, 117, 113, ... .

Here, a = 121 and d = 117 − 121 = −4

Let the n^th term of the given AP be the first negative term. Then,

$$\begin{gathered} a_{n }<0 \\ \Rightarrow 121+\left(n-1\right)\times \left(-4\right)<0 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 125-4n<0 \\ \Rightarrow -4n<-125 \end{gathered}$$
$$\begin{gathered} \Rightarrow n>\frac{125}{4}=31\frac{1}{4} \\ \therefore n=32 \end{gathered}$$

Hence, the 32nd term is the first negative term of the given AP.

Answer:

The given AP is 20, $19\frac{1}{4}$, $18\frac{1}{2}$, $17\frac{3}{4}$, ... .

Here, a = 20 and d = $19\frac{1}{4}-20=\frac{77}{4}-20=\frac{77-80}{4}=-\frac{3}{4}$

Let the n^th term of the given AP be the first negative term. Then,

$$\begin{gathered} a_{n }<0 \\ \Rightarrow 20+\left(n-1\right)\times \left(-\frac{3}{4}\right)<0 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 20+\frac{3}{4}-\frac{3}{4}n<0 \\ \Rightarrow \frac{83}{4}-\frac{3}{4}n<0 \end{gathered}$$
$$\begin{gathered} \Rightarrow -\frac{3}{4}n<-\frac{83}{4} \\ \Rightarrow n>\frac{83}{3}=27\frac{2}{3} \\ \therefore n=28 \end{gathered}$$

Hence, the 28th term is the first negative term of the given AP.

Answer:

We have:
T₇ = a + (n - 1)d 
⇒ a + 6d = -4           ...(1)

​T₁₃ = a + (n - 1)d  
⇒ a + 12d = -16          ...(2)
On solving (1) and (2), we get:
a = 8 and d = -2

Answer:

In the given AP, let the first term be a and the common difference be d.
Then, T_n = a + (n - 1)d ​
Now, T₄ = a + (4 - 1)d 
⇒ a + 3d  = 0        ...(1)

Answer:

$$\begin{gathered} a_6=a+5d=0 \\ \mathrm{And} \\ {a}_{33}=a+32d=a+5d+27d \\ =27d \left(\because a+5d=0\right) .....\left(1\right) \\ 3 \mathrm{times} \mathrm{of} 15\mathrm{th} \mathrm{term} \mathrm{means} \\ 3a_{15}=3\left(a+14d\right)=3a+42d \\ =3a+15d+27d \\ =3\left(a+5d\right)+27d \\ =27d .....\left(2\right) \end{gathered}$$
Thus, (1) = (2)
Hence, 33rd term is three times its 15th term.

Answer:

Let a be the first term and d be the common difference of the AP. Then,

$$\begin{gathered} a_4=11 \\ \Rightarrow a+\left(4-1\right)d=11 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+3d=11 .....\left(1\right) \end{gathered}$$

Now,

$a_5+a_7=34$                (Given)
$$\begin{gathered} \Rightarrow \left(a+4d\right)+\left(a+6d\right)=34 \\ \Rightarrow 2a+10d=34 \\ \Rightarrow a+5d=17 .....\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} 11-3d+5d=17 \\ \Rightarrow 2d=17-11=6 \\ \Rightarrow d=3 \end{gathered}$$

Hence, the common difference of the AP is 3.

Answer:

Let a be the first term and d be the common difference of the AP. Then,

$$\begin{gathered} a_9=-32 \\ \Rightarrow a+\left(9-1\right)d=-32 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+8d=-32 .....\left(1\right) \end{gathered}$$

Now,

$a_{11}+a_{13}=-94$                (Given)
$$\begin{gathered} \Rightarrow \left(a+10d\right)+\left(a+12d\right)=-94 \\ \Rightarrow 2a+22d=-94 \\ \Rightarrow a+11d=-47 .....\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} -32-8d+11d=-47 \\ \Rightarrow 3d=-47+32=-15 \\ \Rightarrow d=-5 \end{gathered}$$

Hence, the common difference of the AP is −5.

Answer:

Let a be the first term and d be the common difference of the AP. Then,

$$\begin{gathered} a_7=-1 \\ \Rightarrow a+\left(7-1\right)d=-1 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+6d=-1 .....\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} a_{16}=17 \\ \Rightarrow a+15d=17 .....\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} -1-6d+15d=17 \\ \Rightarrow 9d=17+1=18 \\ \Rightarrow d=2 \end{gathered}$$

Putting d = 2 in (1), we get

$$\begin{gathered} a+6\times 2=-1 \\ \Rightarrow a=-1-12=-13 \end{gathered}$$

$$\begin{gathered} \therefore a_n=a+\left(n-1\right)d \\ =-13+\left(n-1\right)\times 2 \\ =2n-15 \end{gathered}$$

Hence, the n^th term of the AP is (2n − 15).

Answer:

Let a be the first term and d be the common difference of the AP. Then,

$4\times a_4=18\times a_{18}$      (Given)
$$\begin{gathered} \Rightarrow 4\left(a+3d\right)=18\left(a+17d\right) \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 2\left(a+3d\right)=9\left(a+17d\right) \\ \Rightarrow 2a+6d=9a+153d \\ \Rightarrow 7a=-147d \end{gathered}$$
$$\begin{gathered} \Rightarrow a=-21d \\ \Rightarrow a+21d=0 \\ \Rightarrow a+\left(22-1\right)d=0 \\ \Rightarrow a_{22}=0 \end{gathered}$$       

Hence, the 22nd term of the AP is 0.

Answer:

Let a be the first term and d be the common difference of the AP. Then,

$10\times a_{10}=15\times a_{15}$      (Given)
$$\begin{gathered} \Rightarrow 10\left(a+9d\right)=15\left(a+14d\right) \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 2\left(a+9d\right)=3\left(a+14d\right) \\ \Rightarrow 2a+18d=3a+42d \\ \Rightarrow a=-24d \end{gathered}$$
$$\begin{gathered} \Rightarrow a+24d=0 \\ \Rightarrow a+\left(25-1\right)d=0 \\ \Rightarrow a_{25}=0 \end{gathered}$$       

Hence, the 25th term of the AP is 0.

Answer:

Let the common difference of the AP be d.

First term, a = 5

Now,
$a_1+a_2+a_3+a_4=\frac{1}{2}\left(a_5+a_6+a_7+a_8\right)$               (Given)
$\Rightarrow a+\left(a+d\right)+\left(a+2d\right)+\left(a+3d\right)=\frac{1}{2}\left[\left(a+4d\right)+\left(a+5d\right)+\left(a+6d\right)+\left(a+7d\right)\right]$                 $\left[a_n=a+\left(n-1\right)d\right]$
$$\begin{gathered} \Rightarrow 4a+6d=\frac{1}{2}\left(4a+22d\right) \\ \Rightarrow 8a+12d=4a+22d \\ \Rightarrow 22d-12d=8a-4a \\ \Rightarrow 10d=4a \end{gathered}$$
$$\begin{gathered} \Rightarrow d=\frac{2}{5}a \\ \Rightarrow d=\frac{2}{5}\times 5=2 \left(a=5\right) \end{gathered}$$

Hence, the common difference of the AP is 2.

Answer:

Let a be the first term and d be the common difference of the AP. Then,

$a_2+a_7=30$                     (Given)
$$\begin{gathered} \therefore \left(a+d\right)+\left(a+6d\right)=30 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 2a+7d=30 .....\left(1\right) \end{gathered}$$

Also,
$a_{15}=2a_8-1$                   (Given)
$$\begin{gathered} \Rightarrow a+14d=2\left(a+7d\right)-1 \\ \Rightarrow a+14d=2a+14d-1 \\ \Rightarrow -a=-1 \\ \Rightarrow a=1 \end{gathered}$$

Putting a = 1 in (1), we get
$$\begin{gathered} 2\times 1+7d=30 \\ \Rightarrow 7d=30-2=28 \\ \Rightarrow d=4 \end{gathered}$$

So,
$a_2=a+d=1+4=5$
$a_3=a+2d=1+2\times 4=9$,...

Hence, the AP is 1, 5, 9, 13,...

Answer:

Let the n^th term of the given progressions be t_n and T_n, respectively.
The first AP is 63, 65, 67,...
Let its first term be a and common difference be d.
Then a = 63 and d = (65 - 63) = 2
So, its n_^th term is given by
t_n = a + (n - 1)d 
⇒ 63 + (n - 1) ⨯ 2
⇒ 61 + 2n

The second AP is 3, 10, 17,...
Let its first term be A and common difference be D.
Then A = 3 and D = (10 - 3) = 7
So, its n_^th term is given by
T_n = A + (n - 1)D 
⇒ 3 + (n - 1) ⨯ 7
⇒ 7n - 4
Now, t_n = ​T_n
 ⇒ 61 + 2n​ = 7n - 4​
⇒ 65 = 5n
⇒ n = 13
Hence, the 13^th terms of the AP's are the same.

Answer:

Let a be the first term and d be the common difference of the AP. Then,

$a_{17}=2a_8+5$                (Given)
$$\begin{gathered} \therefore a+16d=2\left(a+7d\right)+5 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+16d=2a+14d+5 \\ \Rightarrow a-2d=-5 .....\left(1\right) \end{gathered}$$

Also,
$a_{11}=43$           (Given)
$\Rightarrow a+10d=43 .....\left(2\right)$

From (1) and (2), we get
$$\begin{gathered} -5+2d+10d=43 \\ \Rightarrow 12d=43+5=48 \\ \Rightarrow d=4 \end{gathered}$$

Puting d = 4 in (1), we get
$$\begin{gathered} a-2\times 4=-5 \\ \Rightarrow a=-5+8=3 \end{gathered}$$

$$\begin{gathered} \therefore a_n=a+\left(n-1\right)d \\ =3+\left(n-1\right)\times 4 \\ =4n-1 \end{gathered}$$

Hence, the n^th term of the AP is (4n − 1).

Answer:

Let a be the first term and d be the common difference of the AP. Then,

$a_{24}=2a_{10}$           (Given)
$$\begin{gathered} \Rightarrow a+23d=2\left(a+9d\right) \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+23d=2a+18d \\ \Rightarrow 2a-a=23d-18d \\ \Rightarrow a=5d .....\left(1\right) \end{gathered}$$

Now,
$$\begin{gathered} \frac{a_{72}}{a_{15}}=\frac{a+71d}{a+14d} \\ \Rightarrow \frac{a_{72}}{a_{15}}=\frac{5d+71d}{5d+14d} \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow \frac{a_{72}}{a_{15}}=\frac{76d}{19d}=4 \\ \Rightarrow a_{72}=4\times a_{15} \end{gathered}$$

Hence, the 72nd term of the AP is 4 times its 15th term.

Answer:

Let a be the first term and d be the common difference of the AP. Then,

$$\begin{gathered} a_{19}=3a_6 \left(\mathrm{Given}\right) \\ \Rightarrow a+18d=3\left(a+5d\right) \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+18d=3a+15d \\ \Rightarrow 3a-a=18d-15d \end{gathered}$$
$\Rightarrow 2a=3d .....\left(1\right)$

Also,
$$\begin{gathered} a_9=19 \left(\mathrm{Given}\right) \\ \Rightarrow a+8d=19 .....\left(2\right) \end{gathered}$$

From (1) and (2), we get
$$\begin{gathered} \frac{3d}{2}+8d=19 \\ \Rightarrow \frac{3d+16d}{2}=19 \\ \Rightarrow 19d=38 \\ \Rightarrow d=2 \end{gathered}$$

Putting d = 2 in (1), we get
$$\begin{gathered} 2a=3\times 2=6 \\ \Rightarrow a=3 \end{gathered}$$

So,
$$\begin{gathered} a_2=a+d=3+2=5 \\ {a}_3=a+2d=3+2\times 2=7,... \end{gathered}$$

Hence, the AP is 3, 5, 7, 9, ... .

Answer:

In the given AP, let the first term be a and the common difference be d.
Then T_n = a + (n - 1)d ​
⇒ Tp = a + (p - 1)d = q               ...(i)
     ​

Answer:

Answer:

The two-digit numbers divisible by 6 are 12, 18, 24, ..., 96.

Clearly, these number are in AP.

Here, a = 12 and d = 18 − 12 = 6

Let this AP contains n terms. Then,

$$\begin{gathered} a_n=96 \\ \Rightarrow 12+\left(n-1\right)\times 6=96 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 6n+6=96 \\ \Rightarrow 6n=96-6=90 \\ \Rightarrow n=15 \end{gathered}$$

Hence, there are 15 two-digit numbers divisible by 6.

Answer:

The two-digit numbers divisible by 3 are 12, 15, 18, ..., 99.

Clearly, these number are in AP.

Here, a = 12 and d = 15 − 12 = 3

Let this AP contains n terms. Then,

$$\begin{gathered} a_n=99 \\ \Rightarrow 12+\left(n-1\right)\times 3=99 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 3n+9=99 \\ \Rightarrow 3n=99-9=90 \end{gathered}$$
$\Rightarrow n=30$

Hence, there are 30 two-digit numbers divisible by 3.

Answer:

The three-digit numbers divisible by 9 are 108, 117, 126, ..., 999.

Clearly, these number are in AP.

Here, a = 108 and d = 117 − 108 = 9

Let this AP contains n terms. Then,

$$\begin{gathered} a_n=999 \\ \Rightarrow 108+\left(n-1\right)\times 9=999 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 9n+99=999 \\ \Rightarrow 9n=999-99=900 \\ \Rightarrow n=100 \end{gathered}$$

Hence, there are 100 three-digit numbers divisible by 9.

Answer:

The numbers which are divisible by both 2 and 5 are divisible by 10 also.

Now, the numbers between 101 and 999 which are divisible 10 are 110, 120, 130, ..., 990.

Clearly, these number are in AP.

Here, a = 110 and d = 120 − 110 = 10

Let this AP contains n terms. Then,

$$\begin{gathered} a_n=990 \\ \Rightarrow 110+\left(n-1\right)\times 10=990 \left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 10n+100=990 \\ \Rightarrow 10n=990-100=890 \\ \Rightarrow n=89 \end{gathered}$$

Hence, there are 89 numbers between 101 and 999 which are divisible by both 2 and 5.

Answer:

The numbers of rose plants in consecutive rows are 43, 41, 39,..., 11.

Answer:

Let the amount of the first prize be ₹a.

Since each prize after the first is ₹200 less than the preceding prize, so the amounts of the four prizes are in AP.

Amount of the second prize = ₹(a − 200)

Amount of the third prize = ₹(a − 2 × 200) = ₹(a − 400)

Amount of the fourth prize = ₹(a − 3 × 200) = ₹(a − 600)

Now,

Total sum of the four prizes = ₹2,800

∴ ₹a + ₹(a − 200) + ₹(a − 400) + ₹(a − 600) = ₹2,800

⇒ 4a − 1200 = 2800

⇒ 4a = 2800 + 1200 = 4000

⇒ a = 1000

∴ Amount of the first prize = ₹1,000

Amount of the second prize = ₹(1000 − 200) = ₹800

Amount of the third prize = ₹(1000 − 400) = ₹600

Amount of the fourth prize = ₹( 1000 − 600) = ₹400

Hence, the value of each of the prizes is ₹1,000, ₹800, ₹600 and ₹400.

Answer:

Numbers between 200 and 500 divisible by 8 are 208, 216, ..., 496.
This forms an AP 208, 216, ..., 496.
So, first term (a) = 208
Common difference (d) = 8
$$\begin{gathered} a_n=a+\left(n-1\right)d=496 \\ \Rightarrow 208+\left(n-1\right)8=496 \\ \Rightarrow \left(n-1\right)8=288 \\ \Rightarrow n-1=36 \\ \Rightarrow n=37 \end{gathered}$$
Thus, there are 37 integers between 200 and 500 which are divisible by 8.

Answer:

It is given that (3k − 2), (4k − 6) and (k + 2) are three consecutive terms of an AP.

∴ (4k − 6) − (3k − 2) = (k + 2) − (4k − 6)

⇒ 4k − 6 − 3k + 2 = k + 2 − 4k + 6

⇒ k − 4 = −3k + 8

⇒ k + 3k = 8 + 4

⇒ 4k = 12

⇒ k = 3

Hence, the value of k is 3.

Answer:

It is given that (5x + 2), (4x − 1) and (x + 2) are in AP.
$$\begin{gathered} \therefore \left(4x-1\right)-\left(5x+2\right)=\left(x+2\right)-\left(4x-1\right) \\ \Rightarrow 4x-1-5x-2=x+2-4x+1 \\ \Rightarrow -x-3=-3x+3 \\ \Rightarrow 3x-x=3+3 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2x=6 \\ \Rightarrow x=3 \end{gathered}$$

Hence, the value of x is 3.

Answer:

It is given that (3y − 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP.
$$\begin{gathered} \therefore \left(3y+5\right)-\left(3y-1\right)=\left(5y+1\right)-\left(3y+5\right) \\ \Rightarrow 3y+5-3y+1=5y+1-3y-5 \\ \Rightarrow 6=2y-4 \\ \Rightarrow 2y=6+4=10 \end{gathered}$$
$\Rightarrow y=5$

Hence, the value of y is 5.

Answer:

Since (x + 2), 2x and (2x + 3) are in AP, we have:
2x - (x+2) = (2x+3) -2x
⇒ x - 2 = 3
⇒ x = 5​
∴ x = 5

Answer:

The given numbers are ${\left(a-b\right)}^2, \left(a^2+b^2\right)$ and ${\left(a+b\right)}^2$.
Now,

$\left(a^2+b^2\right)-{\left(a-b\right)}^2=a^2+b^2-\left(a^2-2ab+b^2\right)=a^2+b^2-a^2+2ab-b^2=2ab$

${\left(a+b\right)}^2-\left(a^2+b^2\right)=a^2+2ab+b^2-a^2-b^2=2ab$

So, $\left(a^2+b^2\right)-{\left(a-b\right)}^2={\left(a+b\right)}^2-\left(a^2+b^2\right)=2ab \left(\mathrm{Constant}\right)$

Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.

Answer:

Let the first three numbers in an arithmetic progression be a − d, a, a + d.

The sum of the first three numbers in an arithmetic progression is 15.

a − d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5

Their product is 105.

$$\begin{gathered} \left(a-d\right)\times a\times \left(a+d\right)=105 \\ \Rightarrow \left(a^2-d^2\right)\times a=105 \\ \Rightarrow \left({\left(5\right)}^2-d^2\right)\times 5=105 \\ \Rightarrow {\left(5\right)}^2-d^2=21 \\ \Rightarrow 25-d^2=21 \\ \Rightarrow d^2=25-21 \\ \Rightarrow d^2=4 \\ \Rightarrow d=2, -2 \\ \mathrm{For} d=2, \\ \mathrm{The} \mathrm{numbers} \mathrm{are} 5-2, 5, 5+2 \\ \mathrm{i}.\mathrm{e}. 3, 5, 7 \\ \mathrm{For} d=-2, \\ \mathrm{The} \mathrm{numbers} \mathrm{are} 5+2, 5, 5-2 \\ \mathrm{i}.\mathrm{e}. 7, 5, 3 \end{gathered}$$

Hence, the three numbers are 3, 5, 7 or 7, 5, 3.

Answer:

Let the required numbers be (a - d), a and (a + d). 
Then (a - d) + a + (a + d) = 3
⇒ 3a = 3
⇒ a = 1​
Also, (a - d).a.(a + d)​ = -35
⇒ a(a² - d²) = -35​  
⇒ 1​.​(1​ - d²) = -35​ ​   ​  
⇒ d² = 36 
⇒ d = $\pm$6

Thus, a = 1 and d = $\pm$6
 Hence, the required numbers are ( -5, 1 and  7) or ( 7, 1 and -5).

Answer:

Let the required parts of 24 be (a - d), a and (a + d) such that they are in AP. 
Then (a - d) + a + (a + d) = 24
⇒ 3a = 24  
⇒ a = 8​
Also, (a - d).a.(a + d)​ = 440
⇒ a(a² - d²) = 440​   
⇒ 8​(64​​ -d²) = 440​    ​  
⇒ d² = 64 - 55 = 9 
⇒ d = $\pm$3
 Thus, a = 8 and d = $\pm$3
 Hence, the required parts of 24 are (5, 8,11) or (11, 8, 5).

Answer:

Let the required terms be (a - d), a and (a + d). 
Then (a - d) + a + (a + d) = 21
⇒ 3a = 21
⇒ a = 7​
Also, (a - d)² + a² + (a + d)²​ = 165
⇒ 3a² + 2d² = 165  
⇒ ​(3 ⨯ 49​​ + 2 d²) = 165   ​  
⇒ 2d² = 165 - 147 = 18 
⇒ d²  = 9​
⇒ d = $\pm$3
 Thus, a = 7 and d = $\pm$3
 Hence, the required terms are ( 4, 7,10) or ( 10, 7, 4).

Answer:

Let the required angles be (a - 15)^o, (a - 5)^o, (a + 5)^o and (a + 15)^o, as the common difference is 10 (given).
Then (a - 15)^o + (a - 5)^o + (a + 5)^o + (a + 15)^o = 360^o
⇒ 4a = 360 
⇒ a = 90

Hence, the required angles of a quadrilateral are $\left(90-15\right)°, \left(90-5\right)°, \left(90+5\right)° \mathrm{and} \left(90+15\right)°; \mathrm{or} 75°, 85°, 95° \mathrm{and} 105°.$

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{four} \mathrm{numbers} \mathrm{be} \\ a-3d,a-d,a+d,a+3d \\ \mathrm{According} \mathrm{to} \mathrm{question}: \\ \left(a-3d\right)+\left(a-d\right)+\left(a+d\right)+\left(a+3d\right)=28 \\ \Rightarrow 4a-4d+4d=28 \\ \Rightarrow a=7 \\ \mathrm{And} \\ {\left(a-3d\right)}^2+{\left(a-d\right)}^2+{\left(a+d\right)}^2+{\left(a+3d\right)}^2=216 \end{gathered}$$

$$\begin{gathered} a^2+9d^2-6ad+a^2+d^2-2ad+a^2+d^2+2ad+a^2+9d^2+6ad=216 \\ \Rightarrow 4a^2+20d^2=216 \\ \Rightarrow 4{\left(7\right)}^2+20d^2=216 \\ \Rightarrow d=\pm 1 \end{gathered}$$
$$\begin{gathered} \mathrm{Four} \mathrm{numbers} \mathrm{are}: \\ d=1,a=7 \\ a-3d,a-d,a+d,a+3d \\ 7-3\left(1\right),7-1,7+1,7+3\left(1\right) \\ 4,6,8,10 \\ \mathrm{And} \\ d=-1,a=7 \\ 10,8,6,4 \end{gathered}$$

Answer:

Let the four parts in AP be (a − 3d), (a − d), (a + d) and (a + 3d). Then,

$$\begin{gathered} \left(a-3d\right)+\left(a-d\right)+\left(a+d\right)+\left(a+3d\right)=32 \\ \Rightarrow 4a=32 \\ \Rightarrow a=8 .....\left(1\right) \end{gathered}$$

Also,
$$\begin{gathered} \left(a-3d\right)\left(a+3d\right):\left(a-d\right)\left(a+d\right)=7:15 \\ \Rightarrow \frac{\left(8-3d\right)\left(8+3d\right)}{\left(8-d\right)\left(8+d\right)}=\frac{7}{15} \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow \frac{64-9d^2}{64-d^2}=\frac{7}{15} \\ \Rightarrow 15\left(64-9d^2\right)=7\left(64-d^2\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow 960-135d^2=448-7d^2 \\ \Rightarrow 135d^2-7d^2=960-448 \\ \Rightarrow 128d^2=512 \\ \Rightarrow d^2=4 \end{gathered}$$
$\Rightarrow d=\pm 2$

When a = 8 and d = 2,
$$\begin{gathered} a-3d=8-3\times 2=8-6=2 \\ a-d=8-2=6 \\ a+d=8+2=10 \\ a+3d=8+3\times 2=8+6=14 \end{gathered}$$

When a = 8 and d = −2,
$$\begin{gathered} a-3d=8-3\times \left(-2\right)=8+6=14 \\ a-d=8-\left(-2\right)=8+2=10 \\ a+d=8-2=6 \\ a+3d=8+3\times \left(-2\right)=8-6=2 \end{gathered}$$

Hence, the four parts are 2, 6, 10 and 14.

Answer:

Let the first three terms of the AP be (a − d), a and (a + d). Then,
$$\begin{gathered} \left(a-d\right)+a+\left(a+d\right)=48 \\ \Rightarrow 3a=48 \\ \Rightarrow a=16 \end{gathered}$$

Now,
$$\begin{gathered} \left(a-d\right)\times a=4\left(a+d\right)+12 \left(\mathrm{Given}\right) \\ \Rightarrow \left(16-d\right)\times 16=4\left(16+d\right)+12 \\ \Rightarrow 256-16d=64+4d+12 \\ \Rightarrow 16d+4d=256-76 \end{gathered}$$
$$\begin{gathered} \Rightarrow 20d=180 \\ \Rightarrow d=9 \end{gathered}$$

When a = 16 and d = 9,
$$\begin{gathered} a-d=16-9=7 \\ a+d=16+9=25 \end{gathered}$$

Hence, the first three terms of the AP are 7, 16 and 25.

Answer:

Let the first three numbers in an arithmetic progression be a − d, a, a + d.

The sum of the first three numbers in an arithmetic progression is 18.

a − d + a + a + d = 18
⇒ 3a = 18
⇒ a = 6

The product of the first and third term is 5 times the common difference.

$$\begin{gathered} \left(a-d\right)\left(a+d\right)=5d \\ \Rightarrow a^2-d^2=5d \\ \Rightarrow {\left(6\right)}^2-d^2=5d \\ \Rightarrow 36-d^2=5d \\ \Rightarrow d^2+5d-36=0 \\ \Rightarrow d^2+9d-4d-36=0 \\ \Rightarrow d\left(d+9\right)-4\left(d+9\right)=0 \\ \Rightarrow \left(d+9\right)\left(d-4\right)=0 \\ \Rightarrow d=4, -9 \\ \mathrm{For} d=4, \\ \mathrm{The} \mathrm{numbers} \mathrm{are} 6-4, 6, 6+4 \\ \mathrm{i}.\mathrm{e}. 2, 6, 10 \\ \mathrm{For} d=-9, \\ \mathrm{The} \mathrm{numbers} \mathrm{are} 6+9, 6, 6-9 \\ \mathrm{i}.\mathrm{e}. 15, 6, -3 \end{gathered}$$

Hence, the three numbers are 2, 6, 10 or 15, 6, −3.

Answer:

(i) The given AP is 2, 7, 12, 17, ... .

Here, a = 2 and d = 7 − 2 = 5

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
$$\begin{gathered} S_{19}=\frac{19}{2}\left[2\times 2+\left(19-1\right)\times 5\right] \\ =\frac{19}{2}\times \left(4+90\right) \\ =\frac{19}{2}\times 94 \\ =893 \end{gathered}$$

(ii) The given AP is 9, 7, 5, 3, ... .

Here, a = 9 and d = 7 − 9 = −2

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
$$\begin{gathered} S_{14}=\frac{14}{2}\left[2\times 9+\left(14-1\right)\times \left(-2\right)\right] \\ =7\times \left(18-26\right) \\ =7\times \left(-8\right) \\ =-56 \end{gathered}$$

(iii) The given AP is −37, −33, −29, ... .

Here, a = −37 and d = −33 − (−37) = −33 + 37 = 4

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
$$\begin{gathered} S_{12}=\frac{12}{2}\left[2\times \left(-37\right)+\left(12-1\right)\times 4\right] \\ =6\times \left(-74+44\right) \\ =6\times \left(-30\right) \\ =-180 \end{gathered}$$

(iv) The given AP is $\frac{1}{15},\frac{1}{12},\frac{1}{10},...$ .

Here, a = $\frac{1}{15}$ and d = $\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}$

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
$$\begin{gathered} S_{11}=\frac{11}{2}\left[2\times \left(\frac{1}{15}\right)+\left(11-1\right)\times \frac{1}{60}\right] \\ =\frac{11}{2}\times \left(\frac{2}{15}+\frac{10}{60}\right) \\ =\frac{11}{2}\times \left(\frac{18}{60}\right) \\ =\frac{33}{20} \end{gathered}$$

(v) The given AP is 0.6, 1.7, 2.8, ... .

Here, a = 0.6 and d = 1.7 − 0.6 = 1.1

Using the formula, $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$, we have
$$\begin{gathered} S_{100}=\frac{100}{2}\left[2\times 0.6+\left(100-1\right)\times 1.1\right] \\ =50\times \left(1.2+108.9\right) \\ =50\times 110.1 \\ =5505 \end{gathered}$$