Let the n^th term be zero.
then a_n = 0
⇒ a + (n - l)d = 0
⇒ 150 + (n - 1)(-3) = 0 1
⇒ 150- 3n + 3 = 0
⇒ -3n = -153
∴ n = 51

(i) Since each section of each class plants, the same number of trees as the class number and there are three sections of each class.
\(\therefore\) Total number of trees planted by the students = 3 [ 1 + 2 + 3 + ... + 12 ]
\(=3\left[ \frac{12}{2}(2\times1+(12-1)\times1) \right]=3[6(2+11)]\)
= 18 x 13 = 234
(ii) In a school, trees are used for beautification, to reduce noise pollution and to reduce air pollution.

Yes,  \(\because\)   Monthly fee for class I = Rs 250
Fee increment for next higher class = Rs 50
\(\therefore\)  Sequence becomes 250, 300, 350, 400, ...
Here, a₂ - a₁ = 300 - 250 = 50, a₃ - a₂
= 350 - 300 = 50, a₄ - a₃ = 400 - 350 = 50
i.e.,  common difference (d) is same everywhere.
So, the given list of numbers forms an A.P.
(i) Option (a) ask your friends to contribute for his fee if you find that your pocket money is less than the amount required.

k = 4

Let Ist term of the A.P = a and common difference = d
Sm = \(\frac{m}{2}[2a+(m-1)d]\)
Sn = \(\frac{n}{2}[2a+(n-1)d]\)
A.T.Q.
\(\frac{{S}_{m}}{{S}_{n}}=\frac{{m}^{2}}{{n}^{2}}\)
\(\Rightarrow \)    \(\frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{{m}^{2}}{{n}^{2}}\)
\(\Rightarrow\)     \(\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{{m}^{2}}{{n}^{2}}\times{n}{m}=\frac{m}{n}\)
Replacing m with 2m - 1 and n with 2n - 1 we get
\(\frac{2a+(2m-1-1)d}{2a+(2n-1-1)d}=\frac{2m-1}{2n-1}\)
\(\Rightarrow\)      \(\frac{2a+2(m-1)d}{2a+2(n-1)d}=\frac{2m-1}{2n-1}\)
\(\Rightarrow\)     \(\frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}\)
\(\Rightarrow\)          \(\frac{{a}_{m}}{{a}_{n}}=\frac{2m-1}{2n-1}\)
Hence proved

Given : S_n = 3n² + 2n.
Let t_n = S_n - S_n-1
= (3n² + 2n) - {3(n - 1)² + 2(n - 1)}
= 3n² + 2n - 3n² + 6n - 3 - 2n + 2
Hence, t_n = 6n - 1.

Here, a = 2, d = 4 - 2 = 2
 a_n = a + (n - 1)d
200 = 2 + (n - 1)2
n = 100
There are 100 terms in the given AP.
Now   S_n = \(\frac{n}{2}\)[a + l]
S₁₀₀ = \(\frac{100}{2}\)[2 + 200] = 10100

Here,   t₈ = \(\frac{{t}_{2}}{2}\Rightarrow\) a + 7d = \(\frac{a+d}{2}\)
\(\Rightarrow\)  2a + 14d = a + d \(\Rightarrow\) a = - 13d (i)
and  t₁₁-\(\frac{{t}_{4}}{3}=1\)
\(\Rightarrow\)  a + 10d - \(\frac{a+3d}{3}=1\)
\(\Rightarrow\frac{3a+30d-a-3d}{3}=1\)
\(\Rightarrow\)  2a + 27d = 3
\(\Rightarrow\)   2 ( - 13d ) + 27d = 3  [ using (i) ]
\(\Rightarrow\)     -26d + 27d = 3
\(\Rightarrow\)   d = 3,
\(\therefore\)  from (i)  a = -13 x 3 = - 39
Therefore,,  t₁₅ = a + 14d
= - 39 + 14 x 3
= - 39 + 42 = 3

Let common difference of two AP be d
Now for Ist AP, Ist term =  A  = 3
and for IInd AP, Ist term = a = 8
(i) | A₂ - a₂ | = | ( A + d ) - ( a + d ) |
= | A - a | = | 3 - 8 |
= | - 5 | = 5
\(\therefore\)  Difference is 5.
(ii) | A₄ - a₄ | = | ( A + 3d ) - ( a + 3d ) |
 = | A - a | = | 3 -8 | = 5
(iii) | A₁₀ - a₁₀ | = | ( A + 9d ) - ( a + 9d ) |
= | A - a |= | 3 - 8 | = 5
(iv)  | A₂₀ - a₂₀ | =| ( A + 19d )- ( a + 19d ) |
= | A  - a |= | 3 -8 |= 5

Fare for 1st km = Rs.15
Fare for 2 km = 15 + 8 = Rs.23
Fare foe 3 km = 15 + 2 x 8 = Rs.31
AP is 15, 23, 31,...........

Let 1st term = a and common difference = d.
a₅ = a + 4d,  a₁₀ = a + 9d
According to the question,   5 x a₅ =  10 x a₁₀
\(\Rightarrow\)  5(a + 4d) = 10(a + 9d)
\(\Rightarrow\) 5a + 20d = 10a + 90d
\(\Rightarrow\)  a = -14d
Now a₁₅ = a + 14d
\(\Rightarrow\) a₁₅ = -14d + 14d = 0

a = 25, d = 50 - 25 = 25, x = 1000
A.T.Q.,         a_k = x
\(\Rightarrow\)    a + (k - 1)d = 1000
\(\Rightarrow\) 25 + (k - 1)25 = 1000
\(\Rightarrow\)  (k - 1)25 = 975
\(\Rightarrow\)   k - 1  = \(\frac{975}{25}\)
\(\Rightarrow\)   k - 1 = 39
\(\Rightarrow\)  k = 40

Let Ist term of AP = a and common difference = d.
Now,    a₆ = -10 \(\Rightarrow\)  a +  5d  = -10                 ..(i)
Also,     a₁₀ = -26  \(\Rightarrow\)    a + 9d = -26           ...(ii)
Subtract (i) from (ii),    
a + 9d = - 26
a + 5d = -10
-  -          +
 4d = -16   \(\Rightarrow\) d = - 4
Substituting in (i), we get
a + 5 x ( -4 ) = - 10  \(\Rightarrow\) a = 10
Now,  a₁₅ = a + 14d = 10 + 14 X  - 4 = - 46

Here S_n = 3n² + 4n
\(\Rightarrow\)   S_n-1 = 3 ( n - 1 )² + 4( n - 1 )
= 3 ( n² - 2n + 1 ) + 4n - 4
= 3n² - 6n + 3 + 4n - 4
= 3n² - 2n - 1
\(\therefore\)  an = S_n - S_n-1
= ( 3n² + 4n ) - ( 3n² - 2n - 1 )
a_n = 6n + 1
Change n to ( n - 1), we get
a_{n -1} =6 ( n - 1) + 1
= 6n - 6 + 1 = 6n - 5
\(\therefore\)   a_n - a₁ = 6n + 1- 6n + 5 = 6
\(\because\)   a_n - a_n-1 is constant i.e., d = 6
\(\therefore\)   Sequence is an AP.

Here a = -6, d = - \({11\over2}\) -(-6)=\({1\over2}\)
Let -25 be the sum of n terms of this AP, where n \(\epsilon\) N.
Using      S_n = \({n\over2}\) [2a+(n-1)d], we have
- 25 = \({n\over2}[2(-6)+(n-1)\left(\frac{1}{2}\right)]\)
\(\Rightarrow\)    - 50 = \(n\left( \frac{n-25}{2} \right)\)  \(\Rightarrow\)  - 100 = n2 - 25n
\(\Rightarrow\)    n² - 25n + 100 = 0  \(\Rightarrow\) ( n - 5 )( n - 20 ) = 0
\(\therefore\)   n = 5, 20.
Both the values of n are natural numbers and therefore, admissible.

Let Ist term of AP be a and common difference be d.
Now, a₉ = 0
\(\Rightarrow\)     a + 8d = 0 \(\Rightarrow\) a = - 8d                              ...(i)
Now,    a₂₉ = a + 28d = -8d + 28d          [Using eq.(i)]
\(\Rightarrow\)   a₂₉ = 20d                                                 ...(ii)
Also,  a₁₉ = a + 18d = -8d + 18d = 10d
\(\Rightarrow\)    2 X a₁₉ = 2 X 10d = 20d                            ...(iii)
From (ii) and (iii), we have        
a₂₉ = 2 X a₁₉

Here     a = 1, d = 1 and n = 31
S_n = \({n\over2}[2a+(n-1)d]\)
\(={31\over2}[2\times1+(31-1)]\)
\(={31\over2}[2+30]={31\over2}\times 32\)
= 496
\(\therefore\)    Piggy bank amount =  Rs 496
Amount spent = Rs 204
Amount left = Rs 100
Total pocket money = Rs 800

(i) Ist instalment = Rs 3425
IInd instalment = Rs 3225
IIIrd instalment = Rs 3025
and so on
Now 3425, 3225, 3025, .... are in AP, with
a = 3425, d = 3225 - 3425 = -200
Now 6th instalment = a₆ = a + 5d = 3425 + 5 X ( - 200 ) = Rs 2425
(ii)  total amnont paid = \({15\over 2}\) ( 2a + 14d )
= \({15\over 2}\) [ 2 x 3425 + 14 x ( - 200 ) ] = \({15\over 2}\) ( 6850 - 2800 )
= \({15 \over 2}\)  ( 4050 ) = Rs 30375