Use \(u_n=a+(n-1)d\).

9th term: \(a+8d=8\).

4th term: \(a+3d=18\).

Subtract: \(5d=8-18=-10\Rightarrow d=-2\).

Then \(a+8(-2)=8\Rightarrow a=24\).

Answer: \(a=24,\ d=-2\).

3rd term: \(a+2d=1\).

6th term: \(a+5d=10\).

Subtract: \(3d=9\Rightarrow d=3\).

Then \(a=1-2(3)=-5\).

4th term: \(a+3d=-5+9=4\).

8th term: \(a+7d=39\).

4th term: \(a+3d=19\).

Subtract: \(4d=20\Rightarrow d=5\).

Then \(a=39-7(5)=4\).

Sum: \(S_{12}=\frac{12}{2}[2a+(12-1)d]=6[8+55]=378\).

Use \(S_n=\frac{n}{2}(a+l)\).

\(38.5=\frac{n}{2}(1+6)=\frac{7n}{2}\).

So \(77=7n\Rightarrow n=11\).

\(u_1=\frac12(7-1)=3\).

\(u_2=\frac12(7-2)=2.5\).

\(u_3=\frac12(7-3)=2\).

\(u_{20}=\frac12(7-20)=-6.5\).

Find early terms from differences:

\(S_1=2\cdot1\cdot4=8\Rightarrow u_1=8\).

\(S_2=2\cdot2\cdot5=20\Rightarrow u_2=S_2-S_1=12\).

\(S_3=2\cdot3\cdot6=36\Rightarrow u_3=36-20=16\).

\(S_4=2\cdot4\cdot7=56\Rightarrow u_4=56-36=20\).

Terms: 8, 12, 16, 20,… common difference 4, so it is an AP.

8th term: \(a+7d\). 4th term: \(a+3d\).

Given \(a+7d=2(a+3d)\Rightarrow a=d\).

20th term: \(a+19d=40\Rightarrow d+19d=40\Rightarrow d=2\).

So \(a=2\).

\(S_{10}=\frac{10}{2}[2a+9d]=5[4+18]=110\).

1st term: \(a\). 3rd: \(a+2d\). 5th: \(a+4d\).

Given \(a+(a+4d)=18\Rightarrow a+2d=9\).

Also \(a+4d=(a+2d)+6\Rightarrow 2d=6\Rightarrow d=3\).

Then \(a=9-2(3)=3\).

\(S_{12}=\frac{12}{2}[2a+11d]=6[6+33]=234\).

Here \(a=2\), \(d=-7\).

\[ S_{30}=\frac{30}{2}[2a+29d] =15[4+29(-7)] =15[4-203] =15(-199) =-2985. \]

\(u_n=S_n-S_{n-1}\).

\(S_1=1-3=-2\Rightarrow u_1=a=-2\).

\(S_2=4-6=-2\Rightarrow u_2=0\).

So \(d=u_2-u_1=0-(-2)=2\).

10th term: \(u_{10}=a+9d=-2+18=16\).

The integers 1 to \(n\) form an AP with \(a=1\), \(l=n\), \(d=1\).

Sum:

\[ S_n=\frac{n}{2}(a+l)=\frac{n}{2}(1+n)=\frac12 n(n+1). \]

Odd numbers: \(21,23,25,\ldots,99\).

\(a=21\), \(l=99\), \(d=2\).

Number of terms:

\[ n=\frac{99-21}{2}+1=39+1=40. \]

Sum:

\[ S_{40}=\frac{40}{2}(21+99)=20\cdot120=2400. \]

20th term: \(a+19d=-41\).

Sum: \(S_{20}=\frac{20}{2}[2a+19d]=-440\Rightarrow 2a+19d=-44\).

Subtract equations: \((2a+19d)-(a+19d)= -44-(-41)\Rightarrow a=-3\).

Then \(-3+19d=-41\Rightarrow 19d=-38\Rightarrow d=-2\).

\[ S_{10}=\frac{10}{2}[2a+9d] =5[2(-3)+9(-2)] =5[-6-18] =-120. \]

Let smallest be \(a\). Largest is \(a+6d\).

Given \[ a+6d=5a \Rightarrow 6d=4a \Rightarrow d=\frac{2}{3}a \]

Sum of 7 terms:

\[ S_7=\frac{7}{2}[2a+6d] =\frac{7}{2}[2a+6(\tfrac{2}{3}a)] =\frac{7}{2}[2a+4a] =\frac{7}{2}\cdot6a =21a. \]

\(21a=630\Rightarrow a=30\).

Largest \(=5a=150\) cm.

This is an AP with \(a=10\), \(d=2\), last term \(l=48\).

\[ l=a+(n-1)d \Rightarrow 48=10+2(n-1) \Rightarrow 38=2(n-1) \Rightarrow n-1=19 \Rightarrow n=20. \]