Let the four consecutive number in AP are
a - 3d, a - d, a + d, a + 3d
Then, we have
(a - 3d) + (a - d) + (a + d) + (a + 3d)
⇒ 4a = 32
⇒ a = 8
Also, it is given that the ratio of the product of first and the last terms to the product of the two middle terms is 7:15, therefore we have
 \(\frac { (a-3d)(a+3d) }{ (a-d)(a+d) } =\frac { 7 }{ 15 } \ \)
\(\ \Rightarrow \ \frac { a^{ 2 }-9d^{ 2 } }{ a^{ 2 }-d^{ 2 } } =\frac { 7 }{ 15 } \Rightarrow 15a^{ 2 }-135d^{ 2 }=7a^{ 2 }-7d^{ 2 }\ \)
\(\ \Rightarrow 8a^{ 2 }-128d^{ 2 }\Rightarrow a^{ 2 }=16d^{ 2 }\)
\(\\ \Rightarrow \ 64=16d^{ 2 }\quad \quad \ d \left[ \because \quad a=8 \right] \)
\(\\ \Rightarrow \ d^{ 2 }=4\Rightarrow d=\pm 2\)
Hence, the number are 2, 6, 10, 14 or 14, 10, 6, 2.

8 n-3; 5,13,21,...

28

(a + 2) + (b + 1),(a + 2) + (b + 2),(a + 3) + (b + 2)

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

Here, first instalment, a = Rs 1000
and increases the instalment every month, d = Rs 100
Number of instalments, n  = 30
Then, list of numbers is 
 \(1000,(1000+100),(1000+2\times100),(1000+3\times100)...\) 
i.e. 1000, 1100, 1200, 1300,... which is an AP. 
Amount paid in 30th instalment,
 \({a}_{30}=1000+(30-1)100 \quad[\because {a}_{n}=a+(n-1)d]\) 
 \(= 1000 + 29 \times 100\) 
= 1000 + 2900 = Rs 3900
Amount paid in 30 instalments,
  \({S}_{30}=\frac{30}{2}[2 \times 1000+(30-1)100]\) 
   \([\because {S}_{n}=\frac {n}{2}=\{2a+(n-1)d\}]\) 
 \(= 15[2000+29 \times 100]=15[2000+2900]\) 
= 15[4900 ]= Rs 7500
Hence, amount of loan still, he has to pay
 = Rs 118000 - Rs 73500 = Rs 4500

a = a, d = b - a and a_n = c
a_n = a + ( n - 1 )d  \(\Rightarrow\) c = a + ( n - 1 ) ( b - a )
\(\Rightarrow\)   \({c-a\over b-a}=n-1\Rightarrow{c-a\over b-a}+1=n\)
\(\Rightarrow\)  \({c-a+b-a \over b-a}=n\Rightarrow{c+b-2a\over b-a}=n\)
\(\therefore\)   S_n = \({n\over 2}[a+{a}_{n}]\)
\(={c+b-2a\over2(b-a)}[a+c]\)
\(={(a+c)(b+v-2a)\over2(b-a)}\)

A.T.Q., S₅ + S₇ = 167
\(\Rightarrow\) \({5\over2}[2a+14d]+{7\over2}[2a+6d]=167\)
\(\Rightarrow\) 5 ( a + 2d ) + 7 ( a + 3d ) = 167
\(\Rightarrow\)  12a + 31d = 167                   ...(i)
and  S₁₀ = 235
\(\Rightarrow\)  \({10\over2}[2a+9d]=235\)
\(\Rightarrow\)  2a + 9d = \({235\over4}=47\)                ...(ii)
Multiplying equation (ii) by 6 and then subtractigfrom (i), we have
12a + 31d = 167
12a + 54d = 282
-     -          -      
     -23d = -115
d= \({-115\over-23}=5\)
\(\therefore\)  From (ii), 2a + 9d = 47
\(\Rightarrow\)   2a + 9 x 5 = 47
\(\Rightarrow\)    2a = 47 - 45
\(\Rightarrow\)   2a = 2 \(\Rightarrow\) a = 1
Hence, S₂₀ = \({20\over2}[2a+19d]\)
= 10 [ 2 x 1 + 19x 5 ]
= 10 [ 2 + 95 ] = 10 x 97 = 970

\(\because\)         S₆ = 36 and S₁₆ = 256.
\(\Rightarrow\)   S₆ = \({6\over2}\) [ 2a + 5d ]
   [ \(\because\) S_n = \({n\over2} [ 2a + ( n - 1)d ]\)
\(\Rightarrow\)    S₆ = 3 ( 2a + 5d )
\(\Rightarrow\)    \({36\over3}\) = 2a + 5d 
\(\Rightarrow\)    12 = 2a + 5d                       ...(i)
and   S₁₆ = \({16\over2}[2a+15d]\)
\(\Rightarrow\)       \({256\over8}=2a+15d\)
\(\Rightarrow\)   32 = 2a + 15d                     ...(ii)
Subtracting (i) and (ii), 2n + 5d = 12
2a + 15d = 32
-    -           -    
     -10d = -20   \(\Rightarrow\)  d = 2
\(\therefore\)  From (i), 12 = 2a + 5(2)
12 - 10 = 2a  \(\Rightarrow\)  2a = 2 \(\Rightarrow\) a = 1
Hence   S₁₀ = \({10\over2}[2a+9d]\)
= 5 ( 2 x 1 +  9 x 2  )
= 5 ( 2 + 18 )
\(\Rightarrow\)   S₁₀ = 5 x 20 = 100.

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.