Since, Jaspal Singh repays his loan of Rs.118000, with first instalment of Rs 1000 and increases each instalment by Rs 100.
\(\therefore \) His instalments are Rs 1000, Rs 1100, Rs 1200, Rs 1300, ... which forms an A.P.
Here, first term is Rs 1000 and common difference is Rs 100.
\(\therefore \) 30th instalment = a₃₀ = a + 29d
 = Rs (1000 + 29 \(\times \) 100)
= Rs (1000 + 2900) = Rs 3900
Amount paid in 30 instalments = S₃₀
\(\Rightarrow \) S₃₀ =  Rs \(\quad \frac { 30 }{ 2 } (2\times 1000+29\times 100)\)
 \(=\ Rs\ 15 (2000+2900)\\ =\ Rs\ 15(4900)\ =\ Rs\ 73500\)
Amount of loan still have to pay
 =  Rs (118000 - 73500)
 =  Rs 44500 

Here, in L.H.S. of the given equation, we have
a = - 4 and d = - 1 - ( - 4 ) = - 1 + 4 = 3 and l = x
\(\therefore\)    - 4 + ( - 1 ) +  2 + ... + x = 437
\(\Rightarrow\)    \({n\over2}(-4+x)=437\)   [ \(\because\) S_n = \({n\over2}(a+l)\) ]
\(\Rightarrow\)     n ( - 4 + x ) = 874                  ...(i)
Also,  n ( - 4 + x ) = 874                  ...(ii)
   [ \(\because\) a_n = a + ( n - 1 )d ]
From (i) and (ii), we have
n ( - 4 - 4 + ( n - 1)d) = 874
\(\Rightarrow\)   - 8n + n ( n - 1 )3 = 874
\(\Rightarrow\)  - 8n + 3n² - 3n - 874 = 0  
\(\Rightarrow\)   3n² - 11n - 874 = 0
\(n={{{11\pm\sqrt{(-11)^{2}-4\times3\times(-874)}}}\over{2\times3}}\)
\(={{11\pm\sqrt{121+10488}}\over{5}}\)
\(={{11\pm103}\over{6}}={{11+103}\over{6}},{{11-103}\over{6}}\)
\(=19,{-{92}\over{6}}\) ( Rejecting )
n = 19
From(ii), we obtain
x = - 4 + ( 19 - 1 )3
x = - 4 + 54
x = 50

As per statement of the question, we have|
32 + 36 + 40 + ... + n terms = 2000
\(\Rightarrow\)      \({n\over 2}[2\times32+(n-1)4]=2000\)
[ \(\because\) Here, a = 32, d = 36 - 32 = 4 and no. of  terms be n ]
\(\Rightarrow\)   32n + 2n² - 2n - 2000 = 0
\(\Rightarrow\)   2n² + 3n - 2000 = 0
\(\Rightarrow\)   n² + 15n - 1000 = 0
\(\Rightarrow\)  n² + 40n - 25n - 1000 = 0
\(\Rightarrow\) ( n - 25 )( n + 40 ) = 0
\(\Rightarrow\) Either n - 25 = 0 or        n + 40 =0
\(\Rightarrow\)    n = 25 or     n = - 40
Rejecting n  = - 40, because no. of terms cannot by -ve.
Hence, Yasmeen save Rs 2000 in 25 months.

Length of spiral made up of thirteen consecutive semi-circles
\(=(\pi \times 0.5+\pi \times 1.0+\pi \times 1.5+\pi \times 2.0+\ldots+\pi \times 6.5)\)
[\(\because\) circumference of semi-circle= \(\pi\)r, where, r is radius of circle]
\(\begin{aligned} & =0.5 \pi[1+2+3+4+\ldots+13] \\ \end{aligned}\)
\(\begin{aligned} & =\pi \times 0.5 \times \frac{13}{2}[2 \times 1+(13-1) \times 1] \end{aligned}\)
[\(\because\) the numbers 1, 2,3, .,.13, forms an AP with a =1, d = 2 - 1 = 1. Also,
\(\left.S_n=\frac{n}{2}\{2 a+(n-1) d\}\right]\)
\(=\frac{22}{7} \times \frac{5}{10} \times \frac{13}{2} \times 14=143 \mathrm{~cm}\)

Number of houses preceding no.X
= X - 1
Sum of numbers of houses preceding  X
= 1 + 2 + 3 + ... + X - 1
Number of houses following X = 49 - X
Sum of numbers on the houses following
X = ( X + 1 ) + ( X + 2 ) + ... + 49
\(={49-X\over2}(X+1+49)\)
\(=\left( 49-X\over2 \right)[X+50]\)
\(={{2450-X-{X}^{2}}\over{2}}\)
Now, \({{{X}^{2}-X}\over{2}}={{2450-{X}-{X}^{2}}\over{2}}\)
\(\Rightarrow\)   2X² = 2450
\(\Rightarrow\)  X² = 1225
X = \(\sqrt{1225}\) = 35
\(\therefore\)     X = 35

Let policeman catches the thief in n minutes.
Distance covered by policeman in Ist minute = 100 m
Distance covered by policeman in IInd minute = 110 m
Distance covered by policeman is IIIrd minute = 120 m
Total Distance covered by policeman in n minutes = 100 + 110 +120 + ... + n terns
These terms form an AP.
a = 100, d = 10
\(\therefore\)  S_n = \({n\over2}[2\times100+(n-1)\times10]\)
\(={n\over 2}[200+10n-10]\)
\(={n\over 2}[190+10n]=95n+{5n}^{2}\)
Also, distance covered by thief  in ( n + 1 ) minutes.
= 100 x ( n + 1)m = 100n + 100
\(\Rightarrow\)     5n² + 95n = 100n + 100
\(\Rightarrow\)  5n² - 5n - 100 = 0
\(\Rightarrow\)   n² - n - 20 = 0
\(\Rightarrow\)   ( n - 5 ) ( n + 4 ) = 0

Let number of sides of polygon be n.
The smallest angle = 120°.
\(\because\)  angle are in AP
\(\therefore\)  a = 120°   and common difference d = 5°
\(\therefore\)  Angles are 120°, 125°, 130°, ... up to n terms 
 Now, sum of all the interior angles
\(={n\over2}[2a+(n-1)d]\)
\(={n\over2}[2\times120+(n-1)5]\)
\(={n\over2}(235+5n)\)                              ....(i)
Also, sum of interior angles of a polygon
= ( n - 2 ) ( 180 )                             ....(ii)
From (i) and (ii), we get
\({n\over2}(235+5n)=(n-2)(180)\)
\(\Rightarrow\)   235n + 5n² = 360n - 720 
\(\Rightarrow\)  5n² - 125n + 720 = 0
\(\Rightarrow\)  5 ( n² - 25n + 144 ) = 0
\(\Rightarrow\)     n² - 25n + 144 = 0
\(\Rightarrow\)    ( n - 9 ) ( n - 16 ) = 0
\(\Rightarrow\)       n = 9 or n = 16
When n = 16,
the sixteenth angle = a + 15d
= 120 + 15 x 5 = 195°
Which cannot be an interior angle of a polygon.
\(\therefore\)  n = 9.

Let the number of days in which work was finished be n.
Number of workers on Ist day = 150
Number of workers on IInd day = 146
Number of workers on IIIrd day = 142 and so on
One day equivalent of all the worker
= 150 + 146 + 142 + ... upto to n workers
\(={n\over2}[2\times150+(n-1)\times(-4)]\)
\(={n\over2}(304-4n)=152n-{2n}^{2}\)        ....(i)
If 150 workers would have worked every day then number of days required to finish the work = ( n - 8 )
\(\therefore\)  One day equivalent of workers = 150 ( n - 8) = 150n - 1200                      ...(ii)
From (i) and (ii), we have
152n - 2n² = 150n - 1200
\(\Rightarrow\)   2n² - 2n - 1200 = 0
\(\Rightarrow\)  n² - n - 600 = .0
\(\Rightarrow\)  ( n - 25 ) ( n + 24 ) = 0
\(\Rightarrow\)  n = - 24, n = 25
\(\therefore\) Work has completed in 25 days.

n = 27
Middle most term = \(\frac { n+1 }{ 2 } =\frac { 28 }{ 2 } =14\)
t₁₃ + t₁₅ = 2 + 2 = 4
t₁₂ + t₁₆ = 4 + 4 = 8
t₁₁ + t₁₇ = 6 + 6 = 12
t₁₀ + t₁₈ = 8 + 8 = 16
t₁ + t₂₇ = 26 + 26 = 52
Hence AP becomes 4,8,12,16,......,52
a = 4, d = 4, a_n = 52, n = 13
\({ S }_{ 13 }=\frac { 13 }{ 2 } \left[ 4+52 \right] =364m\)
and distance covered to collect the books = 364 m
Total distance covered = 364 + 364 = 728 m
Maximum distance she travelled carrying a flag is 26 m

Numbers between 100 and 200 are 101, 102, .... 199
a = 101, d = 1,  a_n = 199
\(\therefore\)   a_n = a + ( n - 1 )d
\(\Rightarrow\)    199 = 101 + ( n - 1 )1
\(\Rightarrow\)    199 = 101 + n - 1   \(\Rightarrow\)    99 = n
and    S_n = \({n\over 2}[a+{a}_{n}]\)
S_n = \({99\over2}[101+199]={99\over2}\times300=14850\)
\(\therefore\)  Sum of numbers not divisible by 9 = 14850 - 1623 = 13167.