\(\frac{13}{3}=\frac{-4}{3}+\frac{(n-1)}{3}\)
Findd a₉ + a₁₀
3

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.

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

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

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 = \({n\over 2}[a+{a}_{n}]\)
\(={c+b-2a\over2(b-a)}[a+c]\)
\(={(a+c)(b+v-2a)\over2(b-a)}\)

\(\because \)   \({{t}_{11} \over{t}_{18}}={2\over3}\Rightarrow{a+10d\over a+17d}={2\over3}\)
\(\Rightarrow\)  3a + 30d = 2a + 34d \(\Rightarrow\) a = 4d
and  \({{t}_{5}\over{t}_{21}}={a+4d\over a+20d}={4d+4d \over 4d+20d}\)
\(={8d\over 24d}={1\over 3}\)
\(\Rightarrow\)    t₅ : t₂₁ = 1 : 3
\(\therefore\)   \({{S}_{5}\over{S}_{21}}={{{5}\over{2}}[2a+4d]\over{{21}\over2}[2a+20d]}={5(2a+4d)\over21(2a+20d)}\)
\(={5(8d+4d)\over21[8d+20d]=}{60d\over588d}\)
\(={30\over294}={15\over147}={5\over49}\)
\(\Rightarrow\)  S₅: S₂₁ = 5:49.

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

Here     a = 1, d = 1 and n = 31
S_n = \({n\over2}[2a+(n-1)d]\)
\(={31\over2}[2\times1+(31-1)]\)
\(={31\over2}[2+30]={31\over2}\times 32\)
= 496
\(\therefore\)    Piggy bank amount =  Rs 496
Amount spent = Rs 204
Amount left = Rs 100
Total pocket money = Rs 800

\(\because\)         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.