(i) We have a₂ – a₁ = 10 – 4 = 6
a₃ – a₂ = 16 – 10 = 6
a₄ – a₃ = 22 – 16 = 6
i.e., a_{k + 1} – a_k is the same every time.
So, the given list of numbers forms an AP with the common difference d = 6.
The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.

Rs.163,  Rs.153,  Rs.143,  Rs.133,......

Common difference, d = a_n+1 - a_n.
= 5 - 11(n + 1) - (5 - 11n)
= 5 - 11n - 11 - 5 + 11n = -11

Given, the penalty for each succeeding day is Rs 50 more than the preceding day, therefore the penalties for the first day,the second day, the third day etc., will form an AP. Here, a= 200,d= 250 - 200 = 50 and n = 30. Clearly, the money required by the contractor to pay as penalty, if he delayed the work by 30 days, will be S₃₀ We know that
\(\begin{aligned} S_n & =\frac{n}{2}[2 a+(n-1) d] \end{aligned}\)
\(\begin{aligned} \therefore \quad S_{30} & =\frac{30}{2}[2 \times 200+(30-1) 50] \\ \end{aligned}\)
= 15 (400 + 1450) = 15 \(\times\) 1850 = 27750
Hence, the contractor has to pay Rs 27750, if he delayed the work.

Here, we have
\(a_{2}-a_{1}=-\frac{1}{2}-\left ( -\frac{1}{2} \right )=-\frac{1}{2}+\frac{1}{2}=0,\)
\(a_{3}-a_{2}=-\frac{1}{2}-\left (-\frac{1}{2} \right )=-\frac{1}{2}+\frac{1}{2}=0,\)
\(a_{4}-a_{3}=-\frac{1}{2}-\left ( -\frac{1}{2} \right )=-\frac{1}{2}+\frac{1}{2}=0\)
and so on.
Since, the difference of any two consecutive terms is same. Therefore, the given list of numbers forms an AP and its common difference (d) is 0.
Now, next three terms of this AP are
a₅ = a₄ + d = -\(\frac{1}{2}\) + 0 = - \(\frac{1}{2}\)
a₆ = a₅ + d = -\(\frac{1}{2}\) + 0 = - \(\frac{1}{2}\)
and a₇ = a₆ + d = -\(\frac{1}{2}\) + 0 = -\(\frac{1}{2}\)

The distance run by the competitor to picks up the first potato, second potato, third potato, fourth potato, .... are respectively 2 \(\times\) 5, 2 \(\times\) (5 + 3), 2 \(\times\) (5 + 3 + 3), 2 \(\times\) (5 + 3 +3 + 3)... i.e. 10, 16, 22, 28, ...
Clearly, it is an AP with first term, a = 10 and common difference, d =16 - 10 = 6.
\(\because\) Sum of first n terms, \(\begin{aligned} & S_n=\frac{n}{2}[2 a+(n-1) d] \\ \end{aligned}\)
\(\therefore\) Sum of first 10 terms, \(\begin{aligned} & S_{10}=\frac{10}{2}[2 \times 10+(10-1) \times 6] \end{aligned}\)
                                                                         [\(\because\) n = 10, given]
= 5 (20 + 54) = 5 \(\times\)74 = 370
Hence, the total distance the competitor has to run is 370 m.

16 rows, 15 bricks.

Given, the sum of first n terms,
  \({ S }_{ n }=4n-{ n }^{ 2 }\)                ...(i)
On putting n = 1 in Eq. (i), we get 
 \({ S }_{ 1 }=4\times 1-{ 1 }^{ 2 }=4-1=3\) 
Thus, sum of first term = 3
On putting n = 2 in Eq. (i), we get
  \({ S }_{ 2 }=4\times 2-{ 2 }^{ 2 }=8-4=4\) 
Thus, sum of first two terms = 4
\(\because\) the n th term of an AP, a_n = S_n - S_n-1
\(\therefore\) Second term = S₂ - S₁ = 4 - 3 = 1
On putting n = 3 in Eq. (i), we get
S₃ = 4 \(\times\) 3 - 3² = 12 - 9 = 3
\(\therefore\) third term = S₃ - S₂ = 3 - 4 = -1
On putting n = 9 in eq. (i), we get
S₉ = 4 \(\times\) 9 - 9² = 36 - 81 = -45
Again, putting n = 10 in Eq. (i), we get
S₁₀ = 4 \(\times\)10 - 10² = 40 - 100 = -60
\(\therefore\) 10th term = S₁₀ - S₉
= -60 - (-45) = -60 + 45 = -15
Now, on replacing n by n - 1 in Eq. (i), we get
\(\begin{aligned} S_{n-1} & =4(n-1)-(n-1)^2 \\ \end{aligned}\)
\(\begin{aligned} & =4 n-4-n^2+2 n-1=-n^2+6 n-5 \\ \end{aligned}\)
\(\begin{aligned} \therefore \quad n \text {th term } & =S_n-S_{n-1} \\ \end{aligned}\)
\(\begin{aligned} & =4 n-n^2-\left(-n^2+6 n-5\right) \end{aligned}\)
\(\begin{aligned} & =4 n-n^2+n^2-6 n+5=5-2 n \end{aligned}\)

Given, S₇ = 49 and S₁₇ = 289.
Let a be the first term and d be the common difference of given AP. Then,
\(\begin{aligned} S_7=49 \Rightarrow \frac{7}{2}[2 a+6 d]=49 \Rightarrow a+3 d=7 \ldots \text { (i) } \\ \end{aligned}\)
and \(\begin{aligned} S_{17}=289 \Rightarrow \frac{17}{2}[2 a+16 d]=289 \end{aligned}\)
\(\Rightarrow\) a + 8d = 17 .....(ii)
On subtracting Eq. (i) from Eq. (i), we get
5d = 10 \(\Rightarrow\) d = 2
On substituting d = 2 in Eq. (i), we get
a + 6 = 7 \(\Rightarrow\)  a = 1
Now, \(S_n=\frac{n}{2}[2 a+(n-1) d]=\frac{n}{2}[2+(n-1) 2]\)
= n[1 + n - 1] = n²

Given AP is \(20, 19 \frac { 1 } { 3 }, 18 \frac { 2 } { 3 },...\)
Here,  a = 20
and    \(d=19\frac { 1 }{ 3 } -20=\frac { 58 }{ 3 } -20=\frac { 58-60 }{ 3 } =\frac { -2 }{ 3 } \) 
Let n terms of given AP be required to get sum 300.
We know that,      
 \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\Rightarrow\)   \(300=\frac { n }{ 2 } \left[ 2\left( 20 \right) +\left( n-1 \right) \left( \frac { -2 }{ 3 } \right) \right] \)
\(\Rightarrow\)   \(600=n\left[ 40-\frac { 2 }{ n } n+\frac { 2 }{ 3 } \right] \)
\(\Rightarrow\)   \(600=\frac { 1 }{ 3 } \left[ 120n-2{ n }^{ 2 }+2n \right] \)
\(\Rightarrow\)    \(600\times 3+2{ n }^{ 2 }-122n=0\)
\(\Rightarrow\)   \({ n }^{ 2 }-61n+900=0\)          [dividing by 2]
\(\Rightarrow\)    \({ n }^{ 2 }-36n-25n+900=0\)
\(\Rightarrow\)    \({ n }^{ 2 }-36n-25n+900=0\)
\(\Rightarrow\)   \(\left( n-36 \right) \left( n-25 \right) =0\)
\(\Rightarrow\)  n = 36 or 25
Since, here a is positive and d is negative, so both values of n are possible. Hence, sum of 25 terms of given AP = Sum of 36 terms of given AP = 300.

Given, initial money P = Rs2000
Rate of interest, R = 7% per yr; Time, T = 1,2,3,4,...
We know that, simple interest is given by the following formula
  \(SI=\frac{PRT}{100}\)
\(\therefore\) SI at the end of 1st year \(= \frac{2000\times7\times1}{100}\) = 140
SI at the end of 2nd year \(= \frac{2000 \times 7 \times 2 } { 100 }\) = Rs 280
SI at the end of 3rd year \(=\frac {2000 \times 7 \times 3} {100 }\)  = Rs 420
Thus, list of numbers is 140, 280, 420,...
Here, 280 - 140 = 420 - 280 = 140
So, above list of numbers is an AP, whose first term (a) = 140 and common difference (d) = 140.
Now, SI at the end of 20 yr will be equal to 20th term of the above AP.
​​​​​​\(\therefore\)        \({a}_{20}=a+(20-1)d\)
 \(=140+19\times 140=140+2660=2800\) 

Let the first term and the common difference of the given AP be a and d, respectively.
According to the question,
Third term + Seventh term = 6
\(\Rightarrow\) (a + 2d) + (a + 6d) = 6
\(\Rightarrow\)  a + 4d = 3  ....(i)
and Third term x Seventh term = 8
\(\Rightarrow\) (a + 2d) (a + 6d) =8
\(\Rightarrow\) {(a + 4d) - 2d}{(a + 4d) + 2d} = 8
\(\Rightarrow\) (3- 2d)(3 + 2d) = 8    [using Eq. (i)]
\(\Rightarrow\) 9 - 4d² = 8  [\(\because\)(a - b) (a + b) = a² - b²]
\(\Rightarrow\) 4d² = 9 - 8
\(\Rightarrow \quad d^2=\frac{1}{4}\)
\(\Rightarrow \quad d= \pm \frac{1}{2}\)
When, d = \(\frac{1}{2}\), then from Eq. (i), we get
\(\begin{aligned} a+4\left(\frac{1}{2}\right) & =3 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & a+2 & =3 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & a & =3-2 \Rightarrow a=1 \end{aligned}\)
Now, sum of first sixteen terms of the AP,
\(\begin{aligned} S_{16} & =\frac{16}{2}[2 a+(16-1) d] \quad\left[\because S_n=\frac{n}{2}\{2 a+(n-1) d\}\right] \\ \end{aligned}\)
\(\begin{aligned} =8[2 a+15 d]=8\left[2(1)+15\left(\frac{1}{2}\right)\right]\left[\because a=1, d=\frac{1}{2}\right] \\ \end{aligned}\)
\(\begin{aligned} =8\left(2+\frac{15}{2}\right)=8\left(\frac{19}{2}\right)=76 \end{aligned}\)
When, d = -\(\frac{1}{2}\), then from Eq. (i), we get
\(a+4\left(-\frac{1}{2}\right)=3 \Rightarrow a-2=3 \Rightarrow a=5\)
Now, sum of first sixteen terms of this AP,
\(\begin{aligned} S_{16} & =\frac{16}{2}[2 a+(16-1) d]\left[\because S_n=\frac{n}{2}\{2 a+(n-1) d\}\right] \\ \end{aligned}\)
\(\begin{aligned} =8(2 a+15 d) \\ \end{aligned}\)
\(\begin{aligned} =8\left[2(5)+15\left(-\frac{1}{2}\right)\right]=8\left(10-\frac{15}{2}\right)=8\left(\frac{5}{2}\right)=20 \end{aligned}\)
Hence, the sum of first sixteen terms, S₁₆ = 20 or 76.

We know that the formula to calculate simple interest is given by
Simple Interest  \(=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
So, the interest at the end of the 1st year = \(Rs \frac{1000 \times 8 \times 1}{100}=Rs 80\)
The interest at the end of the 2nd year = \(Rs \frac{1000 \times 8 \times 2}{100}=Rs 160\)
The interest at the end of the 3rd year = \(RS \frac{1000 \times 8 \times 3}{100}=Rs 240\)
Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.
So, the interest (in Rs) at the end of the 1st, 2nd, 3rd, . . . years, respectively are 80, 160, 240, .
It is an AP as the difference between the consecutive terms in the list is 80, i.e., d = 80. Also, a = 80
So, to find the interest at the end of 30 years, we shall find a₃₀
Now, a₃₀ = a + (30 – 1) d = 80 + 29 x 80 = 2400
So, the interest at the end of 30 years will be Rs. 2400

Here, a = 21, d = 18 – 21 = – 3 and a_n = – 81, and we have to find n.
As a_n = a + ( n – 1) d,
we have – 81 = 21 + (n – 1)(– 3)
– 81 = 24 – 3n
– 105 = – 3n
So, n = 35
Therefore, the 35th term of the given AP is – 81.
Next, we want to know if there is any n for which a_n = 0. If such an n is there, then
21 + (n – 1) (–3) = 0
i.e., 3 (n – 1) = 21
i.e., n = 8
So, the eighth term is 0.

Here a = \(\frac{a-b}{a+b},d=\frac{3a-2b}{a+b}-\frac{a-b}{a+b}\)
\(={2a-b\over a+b}\) and n = 11.
S_n  = \({n\over2}[2a+(n-1)d]\)
\(\Rightarrow\) S₁₁ = \(\frac{11}{2}\left[ 2\left( a-b\over a+b \right)+(11-1)\left( 2a-b \over a+b \right)\right]\)
\(\Rightarrow\)  S₁₁ = \({11\over2}\times2\left[ {a-b\over a+b }+{5(2a-b)\over a+b} \right]\)
\(\Rightarrow\)   S₁₁ = \(11\left[ {a-b+10a-5b\over a+b} \right]\)
\(\Rightarrow\)   S₁₁ = \(11\left[ 11a-6b\over a+b \right]\)

(i) (b) In first day, Veer takes 51 s to complete the 200 m race. But in each day he takes 2 s lesser than the previous days.
Thus, AP series will formed
51, 49, 47, ...
(ii) (c) Since, Veer wants to achieve the race in 31 s. Let Veer takes n days to achieve the target.
\(\therefore\) T_n = a + (n- 1)d
Here, a = 51, d = 49 - 51 = -2
\(\therefore\) 31 = 51 + (n - 1)(-2)
\(\Rightarrow\) (n - 1) 2 = 20
\(\therefore\) (n - 1) = 10
\(\therefore\) n = 11
Hence, he needs minimum 11 days to achieve the goal.
(iii) (b) In an AP series, we get the series of odd terms. Hence, term 30 is not an AP.
(iv) (a) Given, a_n = 2n + 3
\(\therefore\) Common difference = a_n +1 - a_n
= 2(n + 1) + 3 - (2n + 3)
= 2n + 2 + 3 - 2n - 3 = 2
(v) (a) Given, terms 2x, x + 10, 3x + 2 are in AP.
\(\therefore \quad x+10=\frac{2 x+(3 x+2)}{2}\)
\(\Rightarrow\) 2x + 20 = 5x + 2
\(\Rightarrow\) 3x = 18 \(\Rightarrow\) x = 6

(a) (i)
The given AP is 6,12,18...,
The common difference = 12-6 =6.
(b) (ii)
In the given AP, we have:a=d=6
\(\therefore a_{n}=a+(n-1) d=6+(n-1) 6=6+6 n-6=6 n\) 
(c) (ii)
The number of trees planted by the students of all the sections of class
= 8th term of the given AP
= 6n = 6X8=48
(d) (ii)  T
otal number of trees planted by class 12 students
= 6 X 12 = 72
(e) (iii)
3rd term from the end = \((n-3+1) \text { th term }\) 
= (12-3+1) th term = 10th term
= 6X10=60

Let the numbers on the cards be a, a + d, a + Zd, ...
According to question, We have (a + 5d) + (a + 13d) = -76
\(\Rightarrow\) 2a+18d = -76\(\Rightarrow\)a + 9d= -38 ... (1)
And (a + 7d) + (a + 15d) = -96
\(\Rightarrow\) 2a + 22d = -96 \(\Rightarrow\) a + 11d = -48 ...(2)
From (1) and (2), we get
2d= -10 \(\Rightarrow\) d= -5
From (1), a + 9(-5) = -38 \(\Rightarrow\) a = 7
(i) (b): The difference between the numbers on any two consecutive cards = common difference of the A.P.=-5
(ii) (d): Number on first card = a = 7
(iii) (b): Number on 19th card = a + 18d = 7 + 18(-5) = -83
(iv) (a): Number on 23rd card = a + 22d = 7 + 22( -5) = -103
(v) (d): \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\)