Consider 10, 20, 30, 40, ... 490. According to question
490 = 10 + (n - 1) x 10
[\(\because\) a = 10, an = 490]
and \(S_{n}=\frac{n}{2}[2 \times 10+(n-1) \times 10]\)
On solving, we get n = 49.
= 12,250

Let her pocket money be Rs x. Now, she takes Rs 1 on day 1, Rs 2 on day 2, Rs 3 on day 3 and so on till the end of the month, from this money.
i.e.  1 + 2 + 3 + 4 + ..... + 31
which form an AP, in which number of terms is 31 and first term 
\(\therefore\) Sum of first terms = S ₃₁
Sum of n terms,  \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\therefore\)  \({ S }_{ 31 }=\frac { 31 }{ 2 } \left[ 2\times 1+\left( 31-1 \right) \times 1 \right] \)
\(=\frac { 31 }{ 2 } \left( 2+30 \right) \)
\(=\frac { 31\times 32 }{ 2 } \)
\(=31\times 16=496\)
So, Kanika takes Rs 496 till the end of the month from this money.
Also, she spent Rs 204 of her pocket money and found that at the end of the month, she still has Rs 100 with her.
Now, according to the condition
\(\left( x-496 \right) -204=100\)
\(\Rightarrow\) x - 700 = 100
\(\Rightarrow\)   x = Rs 800
Hence, Rs 800 was her pocket money for the month.

Here, 8 and 20 are the first term and common difference of an A.P.
\(\therefore\)  Sn = \({}n\over2\) [ 2(8) + ( n - 1 ) 20 ] = 8n + 10n² - 10n
= 10n² - 2n
- 30 and 8 are the first term and common difference of another A.P.
\(\therefore\)    S 2n = \({2n \over 2}\) [ 2 (-30) + ( 2n - 1 )8 ]
= - 60n +10n² - 8n = n ( - 60 + 16n - 8 )
= 16n² - 68n
As per statement of the question, we have|
16n² - 68n = 10n² - 2n
\(\Rightarrow\) 16n² - 10n² - 68n + 2n = 0
\(\Rightarrow\)  6n² - 66n = 0
\(\Rightarrow\)   6n ( n - 11 ) = 0
\(\Rightarrow\)  Either n  - 11 = 0   or n = 0
\(\Rightarrow\)        n = 11   or    n = 0  ( Rejecting )
We have   n = 11
Hence, value of n is 11.

Let the first three terms of an A.P. be a - d, a, a + d
\(\therefore\)  a - d + a + a + d = 33
\(\Rightarrow\)     3a = 33
a = 11
Now, according to the given condition|
( a - d ) ( a + d ) = a + 29
( 11 - d ) ( 11 + d ) = 11 + 29
\(\Rightarrow\)     121 - d1 = 40
\(\Rightarrow\)    d² = 81
\(\Rightarrow\)  d = \(\pm\) 9
\(\therefore\)  The required A.P is 2, 11, 20, ... or 20, 11, 2, ...

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 = \(11\left[ 11a-6b\over a+b \right]\)