Let the first term be \(a_1\) and the common difference be \(d\).

Using \(a_n=a_1+(n-1)d\), we get:

\[\left\{\begin{aligned}a_1+2d=6\\ a_1+4d=12\end{aligned}\right.\]

Subtract the first equation from the second:

\[2d=6\]

Hence, \(d=3\).

Then \(a_1=6-2\cdot 3=0\).

The sum of the first six terms is:

\[S_6=\frac{6}{2}\bigl(2a_1+5d\bigr)=3(0+15)=45\]

Answer: 45.

Let the first term be \(b_1\) and the common ratio be \(q\).

Using \(b_n=b_1q^{n-1}\), we have:

\[\left\{\begin{aligned}b_1q^2=60\\ b_1q^3=30\end{aligned}\right.\]

Divide the second equation by the first:

\[q=\frac{30}{60}=\frac{1}{2}\]

Then \(b_1\cdot \left(\frac12\right)^2=60\), so \(b_1=240\).

The first five terms are \(240,120,60,30,15\).

Therefore,

\[S_5=240+120+60+30+15=465\]

Answer: 465.

Using \(a_n=a_1+(n-1)d\):

\[a_5=a_1+4d=12\]

Substitute \(d=4\):

\[a_1+16=12\]

Hence, \(a_1=-4\).

Now find the sum of the first seven terms:

\[S_7=\frac{7}{2}\bigl(2a_1+6d\bigr)=\frac{7}{2}(-8+24)=\frac{7}{2}\cdot 16=56\]

Answer: 56.

Using \(b_2=b_1q\), we have:

\[b_1\cdot 2=4\]

Hence, \(b_1=2\).

The sum of the first five terms is:

\[S_5=b_1\frac{q^5-1}{q-1}=2\cdot\frac{2^5-1}{2-1}=2\cdot 31=62\]

Answer: 62.

For an arithmetic progression, \(a_n=a_1+(n-1)d\).

Then:

\[a_5-a_3=(a_1+4d)-(a_1+2d)=2d\]

Given that \(2d=6\), we get \(d=3\).

Now find the tenth term:

\[a_{10}=a_1+9d=20+9\cdot 3=47\]

Answer: 47.

Let the common ratio be \(q\).

Then:

\[b_2=10q,\qquad b_5=10q^4\]

Given that \(b_5=27b_2\), we have

\[10q^4=27\cdot 10q\]

\[q^3=27\]

Hence, \(q=3\).

Therefore,

\[b_5=10\cdot 3^4=10\cdot 81=810\]

Answer: 810.

Let the first term be \(a_1\) and the common difference be \(d\).

Then:

\[\left\{\begin{aligned}a_1+d=20\\ a_1+3d=16\end{aligned}\right.\]

Subtract the first equation from the second:

\[2d=-4\]

Hence, \(d=-2\).

Then \(a_1=20-(-2)=22\).

Find the sum of the first five terms:

\[S_5=\frac{5}{2}\bigl(2a_1+4d\bigr)=\frac{5}{2}(44-8)=\frac{5}{2}\cdot 36=90\]

Answer: 90.

Let the first term be \(b_1\) and the common ratio be \(q\).

Then:

\[\left\{\begin{aligned}b_1q=6\\ b_1q^2=18\end{aligned}\right.\]

Divide the second equation by the first:

\[q=\frac{18}{6}=3\]

Then \(b_1=\frac{6}{3}=2\).

Find the sum of the first four terms:

\[S_4=b_1\frac{q^4-1}{q-1}=2\cdot\frac{3^4-1}{3-1}=2\cdot\frac{80}{2}=80\]

Answer: 80.

Using \(a_n=a_1+(n-1)d\):

\[a_8=a_1+7d=40\]

Substitute \(d=5\):

\[a_1+35=40\]

Hence \(a_1=5\).

Now find the sum of the first seven terms:

\[S_7=\frac{7}{2}\bigl(2a_1+6d\bigr)=\frac{7}{2}(10+30)=\frac{7}{2}\cdot 40=140\]

Answer: 140.

Using \(b_2=b_1q\):

\[b_1\cdot 5=10\]

Hence \(b_1=2\).

Then the sum of the first four terms is:

\[S_4=b_1\frac{q^4-1}{q-1}=2\cdot\frac{5^4-1}{5-1}=2\cdot\frac{624}{4}=312\]

Answer: 312.

For an arithmetic progression:

\[a_6-a_2=(a_1+5d)-(a_1+d)=4d\]

Given that \(4d=16\), we get \(d=4\).

Then the fifth term is:

\[a_5=a_1+4d=8+4\cdot 4=24\]

Answer: 24.

Let the common ratio be \(q\).

Then:

\[b_6=7q^5,\qquad b_3=7q^2\]

According to the condition:

\[7q^5=8\cdot 7q^2\]

\[q^3=8\]

Hence, \(q=2\).

Then the fourth term is:

\[b_4=7\cdot 2^3=56\]

Answer: 56.

As in the similar problem for this progression, we get:

\[\left\{\begin{aligned}a_1+2d=6\\ a_1+4d=12\end{aligned}\right.\]

Hence \(2d=6\), so \(d=3\).

Then \(a_1=6-2\cdot 3=0\).

The sum of the first ten terms is:

\[S_{10}=\frac{10}{2}\bigl(2a_1+9d\bigr)=5(0+27)=135\]

Answer: 135.

Let the first term be \(b_1\) and the common ratio be \(q\).

Then:

\[\left\{\begin{aligned}b_1q=40\\ b_1q^2=20\end{aligned}\right.\]

Divide the second equation by the first:

\[q=\frac{20}{40}=\frac12\]

Then \(b_1=\frac{40}{1/2}=80\).

The first four terms are \(80,40,20,10\).

Therefore,

\[S_4=80+40+20+10=150\]

Answer: 150.

We have:

\[a_5-a_3=(a_1+4d)-(a_1+2d)=2d\]

Given that \(2d=40\), we get \(d=20\).

Now the sixth term is:

\[a_6=a_1+5d=5+5\cdot 20=105\]

Answer: 105.

Let the common ratio be \(q\).

Then:

\[b_2=20q,\qquad b_5=20q^4\]

According to the condition:

\[20q^4=8\cdot 20q\]

\[q^3=8\]

Hence, \(q=2\).

The fourth term is:

\[b_4=20\cdot 2^3=160\]

Answer: 160.