a₁₈ + a₁₉ + a₂₀ = 225
a₃₅ + a₃₆ + a₃₇ = 429
3, 7, 11, 15, ...

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 

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

T_p = a + ( p - 1 )d
\(\Rightarrow\)     \({1\over q}=a+(p-1)d\)                    ...(i)
T_q = a + ( p - 1 )d
\(\Rightarrow\)     \({1\over q}a+(q-1)d\)                         ...(ii)
Subtracting (ii) from (i), we get
\({p-q\over pq}=(p-q)d\Rightarrow d={1\over pq}\)
Putting d = \({1 \over pq}\) in (i), we have
\({1\over q}=a+{(p-q)\over pq}\Rightarrow a={1\over pq}\)
\(\therefore\)     S_pq = \({pq\over2}[2a+(pq-1)d]\)
or,      S_pq = \({pq\over2}\left[ {{2\over pq}+{(pq-1)\over pq}} \right]\)
\(={1\over 2}(pq+1)\)

Given; S₄ = 40 and S₁₄ = 280.
Sum of n terms, S_n = \({n\over2}\) [ 2a + ( n - 1 )d ]
S₄ = \({4\over2}\) [ 2a + ( 4 - 1 )d] = 40
\(\Rightarrow\)      2[ 2a + 3d ] = 40 \(\Rightarrow\)  2a + 3d = 20               ...(i)
Also,     \({14\over2}\) [ 2a + ( 14 - 1 )d] = 280
\(\Rightarrow\)     7 ( 2a + 13d ) = 280  \(\Rightarrow\) 2a + 43d = 40         ....(ii)
Subtarcting (i) from (ii), we get
10d = 20  \(\Rightarrow\) d  = 2
Putting d = 2 in equation (i), we get 
2a + 3 x 2 = 20  \(\Rightarrow\)  2a = 14    \(\Rightarrow\) a = 7
\(\therefore\)       Sn = \({n\over 2}\) [ 2a + ( n - 1 )d ] = \({n\over 2}[2\times7+(n-1)2 ]\)
\(={n\over2}\) [ 14 + 2n - 2 ] = \({n\over 2}\) ( 2n + 2 ) = n ( n + 6 )

We have S_p = q and S_q = p.
On subtracting, we get
2a+ (p + q - 1)d = -2. Now
\(S_{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d]=\left(\frac{p+q}{2}\right)(-2)=-(p+q)\)
- (p + q)

Odd natural numbers are 1,3,5,7,...
Here, a=1,d=2,n=50
Now, \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \)
\({ s }_{ 50 }=\frac { 50 }{ 2 } \left[ 2(1)+(50-1)2 \right] \)
=25[2+(49)2]
=25[2+98]
=25(100)
=2500

13, 17, 21

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

Let a be the first term and d be the common difference of the given AP.
Given, S₇ = 63              ...(i)
Also,  S₇ + sum of next 7 terms =  63 + 161
\(\Rightarrow\)  S₁₄ = 224                ...(ii)
From (i), we get
\({7\over2}[2a+(7-1)d]=63\)
\(\Rightarrow\)    a + 3d = 9              ...(iii)
From (ii), we get
\({14\over2}[2a+(14-1)d]=224\)
\(\Rightarrow\)    2a + 13d = 32         ..(iv) 
Multiplying (iii) by 2 and then subtracting from (iv), we get
( 2a + 13d ) - ( 2a + 6d ) = 32 - 18
\(\Rightarrow\)   7d = 14     \(\Rightarrow\) d = 2
\(\therefore\)    From (iii), we get
a + 6 = 9      \(\Rightarrow\)  a = 3
\(\therefore\)    For AP; a = 3, d = 2,
a₂₈ = a + ( 28 - 1 )d
= 3 + 27 x 2 = 57