Clearly, volume of concrete required to build the I step, II step, III step,.. are respectively,
\(\begin{aligned} \frac{1}{4} \times \frac{1}{2} \times 50, \\ \end{aligned}\)
\(\begin{aligned} & \left(2 \times \frac{1}{4}\right) \times \frac{1}{2} \times 50,\left(3 \times \frac{1}{4}\right) \times \frac{1}{2} \times 50, \ldots, \\ \end{aligned}\)
\(\begin{aligned} \text { i.e. } \quad \frac{50}{8}, 2 \times \frac{50}{8}, 3 \times \frac{50}{8}, \ldots \end{aligned}\)
Now, total volume of concrete,
\(V=\frac{50}{8}+2 \times \frac{50}{8}+3 \times \frac{50}{8}+\ldots=\frac{50}{8}[1+2+3+\ldots]\)
Note that the numbers in the bracket forms an AP with first term (a) = 1, common difference(d) = 2-1=1 and number of terms(n) =15
\(\begin{aligned} \therefore V & =\frac{50}{8} \times \frac{15}{2}[2 \times 1+(15-1) \times 1] \\ \end{aligned}\)
\(\begin{aligned} & {\left[\because S_n=\frac{n}{2}\{2 a+(n-1) d\}\right] } \\ \end{aligned}\)
\(\begin{aligned} =\frac{50}{8} \times \frac{15}{2} \times(2+14)=\frac{25 \times 15}{8} \times 16=750 \mathrm{~m}^3 \end{aligned}\)
Hence, the total volume of concrete required to build the terrace is 750 m³.

According to the question,
Number of rungs =  \(=\frac{\text { Distance between top and bottom rungs }}{\text { Distance between two rungs }}+1\)
\(\begin{aligned} & =\frac{2 \frac{1}{2} \mathrm{~m}}{25 \mathrm{~cm}}+1=\frac{\frac{5}{2} \times 100 \mathrm{~cm}}{25 \mathrm{~cm}}+1 \quad[\because 1 \mathrm{~m}=100 \mathrm{~cm}] \\ \end{aligned}\)
\(\begin{aligned} =\frac{250 \mathrm{~cm}}{25 \mathrm{~cm}}+1=10+1=11 \end{aligned}\)
Hence, there are 11 rungs.
The length of the wood required for the rungs = Sum of length of 11 rungs
\(=\frac{11}{2}(25+45)\)
\(\begin{array}{r} {\left[\begin{array}{r} \because \text { length of rungs forms an AP with first term } \\ (a)=25 \text { and last term }(l)=45 \text { and } S_n=\frac{n}{2}(a+l) \end{array}\right]} \\ \end{array}\)
\(\begin{array}{r} =\frac{11}{2} \times 70=11 \times 35=385 \mathrm{~cm} \end{array}\)
Hence, the length of the wood required for the rungs is 385 cm.

Let a be the first term and d be the common difference of the given AP.
\(\therefore\)  According to question,
a₁ + a₂ + ... + a₁₀ = 210
and  a₃₆ + a₃₇ + .... + a₅₀ = 2565
Consider, a1 + a₂ + ... + a₁₀ = 210
\(\Rightarrow\)    \({10\over 2}[{a}_{1}+{a}_{10}]=210\)
[ Using Sn = \({n\over 2}(a+l)\)where l is the last term ]
\(\Rightarrow\)    5 [ a + a + 9d ] = 210
\(\Rightarrow\)      2a + 9d = 42                     ...(i)
Again consider,
a₃₆ + a₃₇ + ... + a₅₀ = 2565
\(\Rightarrow\)     \({15\over2}[{a}_{36}+{a}_{50}]=2565\)
\(\Rightarrow\)  a + 35d + a + 49d = 171 x 2
\(\Rightarrow\)    2a + 84d = 171 x 2
\(\Rightarrow\)         a + 42d = 171 x 2                   ...(ii)
On multiplying eq.(ii) by 2 and then subtracting from (i), we get
2a + 9d = 42
-2a \(\pm\) 84d = 342
       -75d = -300    \(\Rightarrow\) d = 4
\(\therefore\)  From (ii), a + 42 x 4 = 171
\(\Rightarrow\)   a = 171 - 168 = 3
\(\therefore\)   a = 3 and d = 4.
Hence, AP is 3, 7, 11, 15, .....

\(\because \)   \({{t}_{11} \over{t}_{18}}={2\over3}\Rightarrow{a+10d\over a+17d}={2\over3}\)
\(\Rightarrow\)  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.

The multiples of 4 between 10 and 250 be 12, 16, 20, ..., 248.
Here, a = 12, d = 16 - 12 = 4 and a_n = 248.
From formula, a_n = a + ( n - 1 )d, we get
12 + ( n - 1 )4 = 248
\(\Rightarrow\)  4 ( n - 1) = 248 - 12   \(\Rightarrow\)  4 ( n - 1 ) = 236
\(\Rightarrow\)  n - 1 = \({236 \over 4}\)  =  59   \(\Rightarrow\)  n = 59 + 1 = 60
\(\because\)         S_n = \({n\over2}\) ( a + l )    \(\Rightarrow\)   S₆₀ = \({60\over2}\) ( 12 + 248 ) = 30 x 260 = 7800