T_n = S_n - S_n-1
T_n-1 = S_n-1-S_n-2
S_n- 2S_n-1+S_n-2 = S_n-S_n-1-S_n-1+S_n-2
=(S_n - S_n-1) - (S_n-1-S_n-2)
=T_n - T_n-1 = d.

Let the sum of k terms of A.P. is S_n = 3k² - k
Now kt^h  term of A.P.
= S_n-S_-1
ak = (3k²-k)-[3(k-l)²-(k-l)]
= 3k² - k - [3k²- 6k + 3 - k + 1]
= 3k²-k-3k² + 7k-4
= 6k-4 
first term a = 6 x 1 - 4 = 2

Here, a = 6, d = 6, n = 5
\(\because\) S = n/2[2a+(n-l)d]
S₅ = 5/2[2x6+(5-1)(6)]
= 5/2[12+4x6]
= 5/2[12+24] = 5/2[36]
= 5x18 = 90

a₁ = 3x + y
a₂ = a₁ +d = 3x + Y + x - Y = 4x
a₃ = a₂ + d = 4x + x - Y = 5x-y
a₄ ;:: a₃ + d = 5x - Y + x - Y
= 6x-2y
So, the four terms are 3x + y, 4x, 5x - y and 6x - 2y.

Three consecutive terms k + 2, 4k - 6 and 3k - 2 are in A.P.
\(\therefore\) (4k-6)-(k+2) = (3k-2)-(4k-6)
\(\Rightarrow\) 4k - 6 - k - 2 = 3k - 2 - 4k + 6
\(\Rightarrow\) 3k - 8 = - k + 4
\(\Rightarrow\) 4k=4+8
\(\Rightarrow\) k = 12/4 =3

Given, 5a₅ = 8a₈
⇒ 5(a + 4d) = 8(a + 7d)
⇒ 5a + 20d = 8a + 56d
⇒ 3a+36d=0
⇒ 3(a + 12d) = 0
⇒ a + 12d = 0
∴  a13 = 0

Here, a = 5, d = - 3
\(\because\)  l = a + (n -1)d
\(\because\)- 49 = 5 + (n -1)(- 3)
\(\Rightarrow\)-49 = 5-3n + 3
\(\Rightarrow\) 3n = 49 + 5 + 3
\(\Rightarrow\)n = 57/3 = 19^th term.

12

We know that, sum of first n natural numbers,
\({ S }_{ n }=\frac { n(n+1) }{ 2 } \)
\(\therefore \ { S }_{ 100 }=\frac { 100\left( 100+1 \right) }{ 2 } =\frac { 100\left( 101 \right) }{ 2 } =5050\)

Given, \({ t }_{ n }=\begin{cases} { n }^{ 2 },where\ n\ in\ even \\ { n }^{ 2 }-1,where\ n\ is\ odd \end{cases}\)
For 19th term, i.e. n = 19 which is odd, we take \({ t }_{ n }={ n }^{ 2 }-1={ \left( 19 \right) }^{ 2 }-1=361-1=360\) 

Here, a_n = n²+1
Put n = 1, 2, 3, ..., we obtain
 a₁ = 1²+1 = 2
a₂ = 2²+1 = 5
 a₃ = 3²+1 = 10
Now, list of numbers becomes 2, 5, 10, ......
\(\therefore \) a₂ - a₁ = 5-2 = 3
a₃-a₂ = 10-5 = 5
Since 3 ​​​​\(\neq \) 5 i.e., a₂-a₁\(\neq \) a₃-a₂
Thus, it does not form an A.P.

We have a = 10, d = 10
\(\therefore \)   First term (a₁) = a = 10
Second term (a₂) = a₁ + d = 10+10 = 20
Third term (a₃) = a₂ + d = 20+10 = 30
Fourth term (a₄) = a₃+d = 30+10 = 40
Hence, the required four terms are 10, 20, 30, 40

Given that a _n = 2n+1
Common difference (d) = a_n+1 - a_n
⇒ d = 2(n+1) + 1 - (2n+1)
⇒ d = 2n + 2 + 1 - 2n -1
⇒ d = 2

Here, a = -2 and d = 0
\(\therefore\)   First four terms of the A.P. are -2, -2, -2, -2, ....

Let the three numbers be a - d, a, a + d
 a - d + a + a + d = 27
\(\Rightarrow\)   3a = 27
\(\Rightarrow\)   a = 9

n = 30

Yes

24th

d = 0 , a₅ = 2.6, a₆ = 2.8

Let 1st term = a and common difference = d
\(\because \) a₁₆ = 10
\(\Rightarrow \) a+15d=10  ....(i)
Now S₃₁ = \(\frac { 31 }{ 2 } (2a+30d)\)
= \(\frac { 31 }{ 2 } \times (2a+15d)\)
= 31x10 (using (i))
=310

9th term from th eend of the A.P.5,9,13,.... 181,185 is
9th term of the A.P.185,181,...13,9,5
Here a=185, d=181-185=-4
9=a+8d=185+8x-4
=185-32=153

Here    a = 8, d = 3 -8 = -5
 ​S₂₂ = \(\frac{22}{2}\)[2a + (22 -1)d]
\(\Rightarrow\)    ​​​​​​S₂₂ = 11(16 - 105) = -979

Here  a = 17,  d = 14\(\frac{1}{2}\) - 17 =\(\frac{29}{2}\) - 17 = \(-\frac{5}{2}\)
Let number of terms in AP = n
 a_n = -38
\(\Rightarrow\) a + (n - 1)d = -38
\(\Rightarrow\)  17 + (n - 1) x \((-\frac{5}{2})\) = -38
\(\Rightarrow\)    (n - 1)\((-\frac{5}{2})\) = -55
\(\Rightarrow\)  (n - 1) = -55 x \((-\frac{2}{5})\) = 22
 n = 23

Let a be the first term and d be the common difference of given AP.
Given that the third term of the AP is
a₃ = 16
\(\Rightarrow\) a + 2d = 16 [\(\because\) a_n = a + (n - 1)d] ...(i)
Also, it is given that
7th term of an AP = 12 + 5th term of an AP i.e.
a₇ = 12 + a₅ \(\Rightarrow\) a₇ - a₅ = 12
\(\Rightarrow\) (a + 6d) - (a + 4d) = 12
\(\Rightarrow\) 2d = 12 \(\Rightarrow\) d = 6
On putting d = 6 in Eq. (i), we get
a + 2 \(\times\) 6 = 16
\(\Rightarrow\) a = 16 - 12 = 4
We know that general form of an AP is a, a + d, a + 2d, a + 3d, ... .
Then, the required AP is
4, 3 + 6, 4 + 2 \(\times\) 6, 4 + 3 \(\times\) 6, ... , i.e. 4, 10, 16, 22, .... .

3, 8, 13, 18, …
For this A.P.,
a = 3
d = a₂ − a₁ = 8 − 3 = 5
Let n^th term of this A.P. be 78.
a_n = a + (n − 1) d
78 = 3 + (n − 1) 5
75 = (n − 1) 5
(n − 1) = 15
n = 16
Hence, 16^th term of this A.P. is 78.

1, 3, 9, 27.......
Here,    a₂ -a₁ = 3 - 1 = 2, a₃ - a₂ = 9 - 3 = 6
 a₂ - a₁ \(\neq\) a₃ - a₂
\(\therefore\)  It is not an AP.

a₁ = -5, a₂ = -1
d = a₂ - a₁ = -1 -(- 5 ) = -1 + 5 = 4
So, first term a₁ = -5, common difference, d = 4

Given, a = -2, d = 0
The first four terms of the AP are -2, -2, -2, and -2.

Let I^st prize be Rs x
\(\therefore\) The series in A.P.is
x, x - 20, x - 40, x - 60,......
a = x, d = - 20, S_n =280, n = 4
\({ S }_{ n }=\frac { n }{ 2 } [2a+(n-1)d]\)
\({ S }_{ n }=\frac { n }{ 2 } [2x+3(-20)]\)
280 = 2[2n - 60]
140 = 2n - 60
2n = 200 \(\Rightarrow\) x = 100
the prizes are Rs 100, Rs 80, Rs 60, Rs 40.

\(\frac { 1 }{ 4 } \)

Here \({a}_{n}=4, n=8, \) and \({S}_{n}=192\)
Let d be the common difference of the given AP.
Then,   \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \) 
\(\therefore\)      \(192=\frac { 8 }{ 2 } \left[ 2\times 3+\left( 8-1 \right) d \right] \) 
\(\Rightarrow\)     \(192=4\left( 6+7d \right) \) 
\(\Rightarrow\)     \(48=6+7d\) 
\(\Rightarrow\)      \(7d=48-6=42\)  
\(\Rightarrow\)        \(d=\frac { 42 }{ 7 } =6\) 
Hence,    d = 6

Let a be the first term and d be the common difference of the given AP.
\(\therefore\)  According to question,
a₁ + a₂ + ... + a₁₀ = 210
and  a₃₆ + a₃₇ + .... + a₅₀ = 2565
Consider, a1 + a₂ + ... + a₁₀ = 210
\(\Rightarrow\)    \({10\over 2}[{a}_{1}+{a}_{10}]=210\)
[ Using Sn = \({n\over 2}(a+l)\)where l is the last term ]
\(\Rightarrow\)    5 [ a + a + 9d ] = 210
\(\Rightarrow\)      2a + 9d = 42                     ...(i)
Again consider,
a₃₆ + a₃₇ + ... + a₅₀ = 2565
\(\Rightarrow\)     \({15\over2}[{a}_{36}+{a}_{50}]=2565\)
\(\Rightarrow\)  a + 35d + a + 49d = 171 x 2
\(\Rightarrow\)    2a + 84d = 171 x 2
\(\Rightarrow\)         a + 42d = 171 x 2                   ...(ii)
On multiplying eq.(ii) by 2 and then subtracting from (i), we get
2a + 9d = 42
-2a \(\pm\) 84d = 342
       -75d = -300    \(\Rightarrow\) d = 4
\(\therefore\)  From (ii), a + 42 x 4 = 171
\(\Rightarrow\)   a = 171 - 168 = 3
\(\therefore\)   a = 3 and d = 4.
Hence, AP is 3, 7, 11, 15, .....

Let Ist term of AP be a and common difference be d.
Now, three middle terms of this AP are a₁₀, a₁₁ and a₁₂
A.T.Q.,  a₁₀ + a₁₁ + a₁₂ = 129
\(\Rightarrow\)  ( a + 9d ) + ( a + 10d ) + ( a + 11d ) = 129
\(\Rightarrow\)    3a + 30d = 129
\(\Rightarrow\)     a + 10d = 43  \(\Rightarrow\)  a = 43 - 10d       ..(i)
Also,  last three terms are a₁₉, a₂₀ and a₂₁
\(\therefore\)   a₁₉ + a₂₀ + a₂₁ = 237
\(\Rightarrow \)  ( a + 18d ) + ( a + 19d ) + ( a + 20d )= 237
\(\Rightarrow\)    3a + 57d = 237   \(\Rightarrow\)  a + 19d = 79
\(\Rightarrow\)  43 - 10d + 19d = 79               [ Using eq.(i) ]
\(\Rightarrow\)  9d = 36  \(\Rightarrow\)  d = 4
When d = 4, equation (i) becomes
a = 43 - 10 x 4 = 3
\(\therefore\)   AP is 3, 7, 11, 15, ...

The given numbers are 7, 10\(\frac{1}{2}\), 14, ....84
\(\because 10 \frac{1}{2}-7=14-10 \frac{1}{2}=\ldots=\frac{7}{2}\)
\(\therefore\) The given numbers forms an AP.
Here, first term, a = 7,
common difference, \(d=10 \frac{1}{2}-7=3 \frac{1}{2}=\frac{7}{2}\)
and last term, l = a_n = 84
\(\because\) a_n = a + (n - 1) d
\(\therefore \quad 84=7+(n-1) \frac{7}{2} \quad\left[\because a=7 \text { and } d=\frac{7}{2}\right]\)
\(\Rightarrow\) n - 1 = 22 \(\Rightarrow\) n = 23
\(\because\) Sum of n terms of an AP,
\(\begin{aligned} S_n=\frac{n}{2}(a+l) & \\ \end{aligned}\)
\(​​\begin{aligned} \therefore \text { Sum of } 23 \operatorname{terms}\left(S_{23}\right) & =\frac{23}{2}(7+84)=\frac{23}{2} \times 91 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{2093}{2}=1046 \frac{1}{2} \end{aligned}\)

Here, a = 100,d = -2, t_n = -10 
Using, t_n = a + (n -1)d
\(\Rightarrow\) -10 = 100 + (n -1)(-2)
\(\Rightarrow\) -10 = l00 - 2n + 2
\(\Rightarrow\) 2n = 112
\(\Rightarrow\) n = 56
\(\therefore\) Here 56^th term is - 10
\(\therefore\) Number of terms in A.P. are 56
\( \therefore \quad { S }_{ n }=\frac { n }{ 2 } (a+l)\)
\({ S }_{ 56 }=\frac { 56 }{ 2 } (100-10)\)
\(=\frac { 56 }{ 2 } (90)\)
= 56 \(\times\) 45
\(\Rightarrow\)  S_n = 2520.

Given, total amount Sandeep had=RS.12000
In the beginning of every month, he spend=RS.500
So, in the beginning of 1st month, he had amount, t₁=RS.12000
In the beginning of 2nd month, he had amount, t₂=12000-500=RS.11500
In the beginning of 3rd  month, he had amount, t₃=1500-500=RS.11000
In the beginning of 4th month, he had amount, t₄=11000-500=RS.10500 and so on.
Now, the list of amount is 12000, 11500, 11000, 10500, ...Here, t₂ - t₁ = t₃ - t₂ = t₄ - t₃ =-500
i.e. t_k+1-t_k is the same everytime.
So, the above list of numbers forms an AP.

Let a be the first term and d be the common difference of the AP, whose 7th term is -1 and 16th term is 17.
Thus, \(a_7=-1\) and \(a_16=17\)
\(\Rightarrow\)    a + (7 - 1) d = -1   \([\because a_n=a+(n-1)d]\)
\(\Rightarrow \)     a + 6d =- 1           ...(i)
and  a + ( 16 - 1 )d = 17
\(\Rightarrow\)  a + 15d = 17  ...(ii)
On solving Eqs. (i) and (ii),
we get a = -1 3 and d = 2
Hence, general term   a_n = a + ( n - d )
 = - 13 + (n - 1) 2
 = 2n - 15
and AP is a ,a + d, a + 2d ,a + 3d,...
 -13,-13 + 2,- 13 + 2(2),- 13 + 3(2),...
i.e.   -13,-11,-9,-7,....

111222333

Multiples of 5 lying between 101 and 999 are 105, 110, 115,......., 999 which are in AP.
Here  a = 105 and d = 5.
a_n = a + (n - 1)d
 995 = 105 + n(n -1)5
 890 = 5n - 5
 895 = 5n
 n = 179
S_n = \(\frac{n}{2}\)[a + l] = \(\frac{179}{2}\)[105 + 995] = \(\frac{179}{2}\) x 1100 = 98450.