The first term \(a = 6\) and the second term is \(\frac{6^2}{6+2} = \frac{36}{8} = \frac{9}{2}\).

The common difference \(d\) is given by:

\(d = \frac{9}{2} - 6 = \frac{9}{2} - \frac{12}{2} = -\frac{3}{2}\)

The sum of the first \(n\) terms of an arithmetic progression is:

\(S_n = \frac{n}{2} \left( 2a + (n-1)d \right)\)

Substituting the values, we have:

\(S_n = \frac{n}{2} \left( 12 + (n-1)(-\frac{3}{2}) \right)\)

We need \(S_n < -480\):

\(\frac{n}{2} \left( 12 - \frac{3}{2}(n-1) \right) < -480\)

\(n \left( 12 - \frac{3}{2}n + \frac{3}{2} \right) < -960\)

\(n \left( \frac{27}{2} - \frac{3}{2}n \right) < -960\)

\(27n - 3n^2 < -1920\)

\(3n^2 - 27n - 1920 > 0\)

Solving the quadratic inequality:

\(n = \frac{9 \pm \sqrt{81 + 2560}}{2}\)

\(n = \frac{9 \pm \sqrt{2641}}{2}\)

Calculating the roots, we find the least integer \(n\) is 31.

The sum of the first \(n\) terms of an arithmetic progression is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\).

Equating this to \(n^2 + 4n\), we have:

\(\frac{n}{2} (2a + (n-1)d) = n^2 + 4n\).

Comparing coefficients, we get:

\(\frac{d}{2} = 1\) and \(a - \frac{1}{2}d = 4\).

Solving these, \(d = 2\) and \(a = 5\).

The \(k\)th term is given by \(a + (k-1)d\).

Substituting the values of \(a\) and \(d\), we have:

\(5 + 2(k-1) > 200\).

Simplifying, \(2k + 3 > 200\).

Solving for \(k\), we get \(k > 98.5\).

Thus, the smallest integer value of \(k\) is \(99\).

The nth term of the arithmetic progression is given by \(a_n = \frac{1}{2}(3n - 15)\).

For the first term \(n = 1\), \(a_1 = \frac{1}{2}(3 \times 1 - 15) = -6\).

For the second term \(n = 2\), \(a_2 = \frac{1}{2}(3 \times 2 - 15) = -4.5\).

The first term \(a = -6\) and the common difference \(d = -4.5 - (-6) = 1.5\).

The sum of the first n terms \(S_n\) of an arithmetic progression is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

Substitute \(S_n = 84\), \(a = -6\), and \(d = 1.5\):

\(84 = \frac{n}{2} (2(-6) + (n-1)1.5)\)

\(84 = \frac{n}{2} (-12 + 1.5n - 1.5)\)

\(84 = \frac{n}{2} (1.5n - 13.5)\)

\(168 = n(1.5n - 13.5)\)

\(168 = 1.5n^2 - 13.5n\)

\(1.5n^2 - 13.5n - 168 = 0\)

Divide the entire equation by 1.5:

\(n^2 - 9n - 112 = 0\)

Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -9\), \(c = -112\):

\(n = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \times 1 \times (-112)}}{2 \times 1}\)

\(n = \frac{9 \pm \sqrt{81 + 448}}{2}\)

\(n = \frac{9 \pm \sqrt{529}}{2}\)

\(n = \frac{9 \pm 23}{2}\)

\(n = 16\) or \(n = -7\)

Since n must be positive, \(n = 16\).

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

The sum of the first \(n\) terms of an arithmetic progression is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

For the first nine terms:

\(S_9 = \frac{9}{2} (2a + 8d) = 117\)

Solving for \(2a + 8d\):

\(9(2a + 8d) = 234\)

\(2a + 8d = 26\) (Equation 1)

The sum of the next four terms (terms 10 to 13) is 91:

\(S_{13} - S_9 = 91\)

\(S_{13} = \frac{13}{2} (2a + 12d)\)

\(\frac{13}{2} (2a + 12d) - \frac{9}{2} (2a + 8d) = 91\)

\(13(2a + 12d) - 9(2a + 8d) = 182\)

\(26a + 156d - 18a - 72d = 182\)

\(8a + 84d = 182\)

\(4a + 42d = 91\) (Equation 2)

Solving Equations 1 and 2 simultaneously:

From Equation 1: \(2a + 8d = 26\)

From Equation 2: \(4a + 42d = 91\)

Multiply Equation 1 by 2:

\(4a + 16d = 52\)

Subtract from Equation 2:

\((4a + 42d) - (4a + 16d) = 91 - 52\)

\(26d = 39\)

\(d = 1.5\)

Substitute \(d = 1.5\) into Equation 1:

\(2a + 8(1.5) = 26\)

\(2a + 12 = 26\)

\(2a = 14\)

\(a = 7\)

Thus, the first term is 7, and the common difference is 1.5.

(i) The distance run on the last day is the 21st term of an arithmetic sequence where the first term is 13 km and the common difference is 1.2 km. The formula for the nth term of an arithmetic sequence is:

\(a_n = a + (n-1) imes d\)

Substituting the given values:

\(a_{21} = 13 + 20 imes 1.2 = 37\) km

(ii) The total distance run over 21 days is the sum of the arithmetic sequence. The formula for the sum of the first n terms is:

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

where \(l\) is the last term. Substituting the given values:

\(S_{21} = \frac{21}{2} \times (13 + 37) = 525\) km

(i) Let the common difference be \(d\). The sum of the first ten terms \(S_{10}\) is given by:

\(S_{10} = \frac{10}{2} (2a + 9d) = 5(2a + 9d)\)

The sum of the next five terms \(S_{11 \text{ to } 15}\) is:

\(S_{11 \text{ to } 15} = \frac{5}{2} [(a + 10d) + (a + 14d)] = 5(a + 12d)\)

Given \(S_{10} = S_{11 \text{ to } 15}\), we have:

\(5(2a + 9d) = 5(a + 12d)\)

\(2a + 9d = a + 12d\)

\(a = 3d\)

Thus, \(d = \frac{a}{3}\).

(ii) The tenth term \(a + 9d\) is 36 more than the fourth term \(a + 3d\):

\(a + 9d = (a + 3d) + 36\)

\(6d = 36\)

\(d = 6\)

Since \(d = \frac{a}{3}\), we have:

\(\frac{a}{3} = 6\)

\(a = 18\)

The first term of the arithmetic progression is a = p and the second term is a + d = 2p, so the common difference d = p.

The nth term of an arithmetic progression is given by:

\(a_n = a + (n-1)d\)

Substituting the known values:

\(p + (n-1)p = 336\)

\(np = 336\)

The sum of the first n terms is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

Substituting the known values:

\(\frac{n}{2} (2p + (n-1)p) = 7224\)

\(\frac{n}{2} (p + np) = 7224\)

Substituting \(np = 336\) into the sum equation:

\(\frac{n}{2} (p + 336) = 7224\)

Solving for n and p:

\(n(p + 336) = 14448\)

Using \(np = 336\), we have:

\(n = \frac{336}{p}\)

Substitute into the sum equation:

\(\frac{336}{p} (p + 336) = 14448\)

Solving gives:

\(336 + \frac{336^2}{p} = 14448\)

\(\frac{336^2}{p} = 14448 - 336\)

\(\frac{336^2}{p} = 14112\)

\(p = \frac{336^2}{14112}\)

\(p = 8\)

Substitute back to find n:

\(n = \frac{336}{8} = 42\)

\(Thus, the values are n = 42 and p = 8.\)

The formula for the nth term of an arithmetic progression is given by:

\(a + (n-1) imes 3 = 94\)

For the sum of the first n terms, we use:

\(\frac{n}{2} [2a + (n-1) imes 3] = 1420\)

Substituting the first equation into the second:

\(\frac{n}{2} [a + 94] = 1420\)

Attempt elimination of a or n:

\(3n^2 - 191n + 2840 = 0\)

Solving this quadratic equation gives:

\(n = 40\)

Substitute back to find a:

\(a + 39 imes 3 = 94\)

\(a + 117 = 94\)

\(a = -23\)

The first term of the arithmetic progression is given as \(a = 6\).

The second term is \(2\), so the common difference \(d\) is \(2 - 6 = -4\).

The formula for the sum of the first \(n\) terms of an arithmetic progression is:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

Substitute \(n = 80\), \(a = 6\), and \(d = -4\) into the formula:

\(S_{80} = \frac{80}{2} (2 \times 6 + 79 \times (-4))\)

\(S_{80} = 40 (12 - 316)\)

\(S_{80} = 40 \times (-304)\)

\(S_{80} = -12,160\)

(i) The first term of the progression is \(p + q\) and the last term is \(p + qn\). The sum of the first n terms, \(S_n\), is given by the formula for the sum of an arithmetic series:

\(S_n = \frac{n}{2} \left( 2(p + q) + (n-1)q \right)\)

or equivalently,

\(S_n = \frac{n}{2} (2p + q + nq)\)

(ii) Using the expression for \(S_n\) from part (i), we have:

\(2(2p + q + 4q) = 40\)

\(3(2p + q + 6q) = 72\)

Solving these equations simultaneously:

From the first equation: \(2(2p + 5q) = 40\) gives \(2p + 5q = 20\).

From the second equation: \(3(2p + 7q) = 72\) gives \(2p + 7q = 24\).

Subtract the first equation from the second:

\((2p + 7q) - (2p + 5q) = 24 - 20\)

\(2q = 4\)

\(q = 2\)

Substitute \(q = 2\) back into \(2p + 5q = 20\):

\(2p + 5(2) = 20\)

\(2p + 10 = 20\)

\(2p = 10\)

\(p = 5\)

Thus, \(p = 5\) and \(q = 2\).

The sum of the first \(n\) terms of an arithmetic progression is given by:

\(S_n = \frac{n}{2} [2a + (n-1)d]\)

Substituting the given values, \(a = -12\) and \(d = 6\), we have:

\(S_n = \frac{n}{2} [-24 + (n-1)6]\)

We need \(S_n > 3000\):

\(\frac{n}{2} [-24 + 6n - 6] > 3000\)

\(\frac{n}{2} [6n - 30] > 3000\)

\(3n(2n - 5) > 3000\)

\(3n^2 - 15n - 3000 > 0\)

Solving the quadratic inequality \(3n^2 - 15n - 3000 = 0\), we find the roots:

\(n = 34.2 \text{ and } n = -29.2\)

Since \(n\) must be a positive integer, the least possible value of \(n\) is \(35\).

The circumference measurements form an arithmetic progression with the first term \(a = 5.00\) m and the second term \(a + d = 5.02\) m, where \(d\) is the common difference.

Thus, \(d = 5.02 - 5.00 = 0.02\) m.

The formula for the \(n\)-th term of an arithmetic progression is \(a_n = a + (n-1) imes d\).

We need to find the circumference 20 years after the first measurement, which is the 21st term (since the first measurement is at year 0).

\(a_{21} = 5.00 + (21-1) imes 0.02\)

\(a_{21} = 5.00 + 20 imes 0.02\)

\(a_{21} = 5.00 + 0.40\)

\(a_{21} = 5.40\) m

Given the sum of the first n terms \(S_n = \frac{1}{2}n(3n + 7)\).

For an arithmetic progression, the sum of the first n terms is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

Equating the two expressions for \(S_n\):

\(\frac{n}{2} (2a + (n-1)d) = \frac{1}{2}n(3n + 7)\)

Cancel \(\frac{n}{2}\) from both sides:

\(2a + (n-1)d = 3n + 7\)

Substitute \(n = 1\):

\(2a = 3(1) + 7\)

\(2a = 10\)

\(a = 5\)

Substitute \(n = 2\):

\(2a + d = 3(2) + 7\)

\(2(5) + d = 13\)

\(10 + d = 13\)

\(d = 3\)

Thus, the first term is 5 and the common difference is 3.

\(The first arithmetic progression (AP1) has first term 15 and common difference 4 (since 19 - 15 = 4).\)

The sum of the first n terms of an arithmetic progression is given by:

\(S_n = \frac{n}{2} [2a + (n-1)d]\)

For AP1, \(a = 15\) and \(d = 4\), so:

\(S_{n1} = \frac{n}{2} [2 \times 15 + (n-1) \times 4]\)

\(S_{n1} = \frac{n}{2} [30 + 4n - 4]\)

\(S_{n1} = \frac{n}{2} [4n + 26]\)

\(The second arithmetic progression (AP2) has first term 420 and common difference -5 (since 415 - 420 = -5).\)

For AP2, \(a = 420\) and \(d = -5\), so:

\(S_{n2} = \frac{n}{2} [2 \times 420 + (n-1) \times (-5)]\)

\(S_{n2} = \frac{n}{2} [840 - 5n + 5]\)

\(S_{n2} = \frac{n}{2} [845 - 5n]\)

Since the sums are equal, \(S_{n1} = S_{n2}\):

\(\frac{n}{2} [4n + 26] = \frac{n}{2} [845 - 5n]\)

Cancel \(\frac{n}{2}\) from both sides:

\(4n + 26 = 845 - 5n\)

\(9n = 819\)

\(n = 91\)

The first term of the arithmetic progression is 16, and the second term is 24. Therefore, the common difference is given by:

\(d = 24 - 16 = 8\)

The sum of the first \(n\) terms of an arithmetic progression is given by:

\(S_n = \frac{n}{2} \left( 2a + (n-1)d \right)\)

Substituting the known values \(a = 16\) and \(d = 8\), we have:

\(S_n = \frac{n}{2} \left( 32 + (n-1) \times 8 \right)\)

We need \(S_n > 20,000\):

\(\frac{n}{2} \left( 32 + 8n - 8 \right) > 20,000\)

\(\frac{n}{2} (24 + 8n) > 20,000\)

\(n(12 + 4n) > 20,000\)

\(4n^2 + 12n - 20,000 > 0\)

Dividing the entire inequality by 4:

\(n^2 + 3n - 5,000 > 0\)

Solving the quadratic equation \(n^2 + 3n - 5,000 = 0\) using the quadratic formula:

\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 1\), \(b = 3\), \(c = -5,000\):

\(n = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-5,000)}}{2 \times 1}\)

\(n = \frac{-3 \pm \sqrt{9 + 20,000}}{2}\)

\(n = \frac{-3 \pm \sqrt{20,009}}{2}\)

Approximating the square root:

\(n \approx \frac{-3 + 141.42}{2}\)

\(n \approx 69.21\)

Since \(n\) must be an integer, we round up to the nearest whole number:

\(n = 70\)

Therefore, 70 terms are needed for the sum to exceed 20,000.

Let the first term be \(a = 32\) and the common difference be \(d\).

The 5th term is given by \(a + 4d = 22\).

Substituting \(a = 32\), we have:

\(32 + 4d = 22\)

\(4d = 22 - 32\)

\(4d = -10\)

\(d = -2.5\)

The last term is \(-28\), so:

\(a + (n-1)d = -28\)

\(32 + (n-1)(-2.5) = -28\)

\(32 - 2.5(n-1) = -28\)

\(32 - 2.5n + 2.5 = -28\)

\(34.5 - 2.5n = -28\)

\(-2.5n = -28 - 34.5\)

\(-2.5n = -62.5\)

\(n = 25\)

The sum of an arithmetic progression is given by:

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

where \(l\) is the last term.

\(S_{25} = \frac{25}{2} (32 + (-28))\)

\(S_{25} = \frac{25}{2} (4)\)

\(S_{25} = 25 \times 2\)

\(S_{25} = 50\)

(i) The distance cycled on the nth day is given by the formula for the nth term of an arithmetic sequence:

\(a_n = a + (n-1) imes d\)

where \(a = 200\) km (the distance on the first day) and \(d = -5\) km (the reduction each day). For May 15th, \(n = 15\):

\(a_{15} = 200 + (15-1) imes (-5)\)

\(a_{15} = 200 - 70 = 130\) km

(ii) The total distance cycled is the sum of an arithmetic series:

\(S_n = \frac{n}{2} [2a + (n-1) imes d]\)

We need \(S_n = 3050\) km, \(a = 200\), and \(d = -5\):

\(\frac{n}{2} [400 + (n-1) imes (-5)] = 3050\)

\(\frac{n}{2} [400 - 5n + 5] = 3050\)

\(n(405 - 5n) = 6100\)

\(5n^2 - 405n + 6100 = 0\)

Solving this quadratic equation gives \(n = 20\).

Therefore, he finishes on May 20th.

(a) The amount of water lost each day forms an arithmetic progression (AP) with the first term, \(a = 10\), and the common difference, \(d = 2\).

The formula for the nth term of an AP is \(a + (n-1)d\).

For the 30th day, \(n = 30\), so the amount lost is:

\(10 + (30-1) imes 2 = 10 + 29 imes 2 = 10 + 58 = 68\) litres.

(b) The total amount of water lost by the nth day is given by the sum of the first n terms of the AP:

\(S_n = \frac{n}{2} (2a + (n-1)d)\).

We need \(S_n = 2000\):

\(\frac{n}{2} (20 + 2(n-1)) = 2000\)

\(n(20 + 2n - 2) = 4000\)

\(n(18 + 2n) = 4000\)

\(2n^2 + 18n - 4000 = 0\)

Solving this quadratic equation gives \(n = 41\).

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

The 12th term is given by:

\(a + 11d = 17\)

The sum of the first 31 terms is given by:

\(\frac{31}{2} (2a + 30d) = 1023\)

Solving the first equation for \(a\):

\(a = 17 - 11d\)

Substitute \(a\) in the second equation:

\(\frac{31}{2} (2(17 - 11d) + 30d) = 1023\)

\(\frac{31}{2} (34 - 22d + 30d) = 1023\)

\(\frac{31}{2} (34 + 8d) = 1023\)

\(31(34 + 8d) = 2046\)

\(34 + 8d = 66\)

\(8d = 32\)

\(d = 4\)

Substitute \(d = 4\) back into \(a = 17 - 11d\):

\(a = 17 - 44 = -27\)

The 31st term is given by:

\(a + 30d = -27 + 30 \times 4 = 93\)

Given that the progression is arithmetic, the common difference \(d\) is 12.

The first term is \(4x\) and the second term is \(x^2\).

Therefore, the common difference equation is:

\(x^2 - 4x = 12\)

Rearranging gives:

\(x^2 - 4x - 12 = 0\)

Solving this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), \(c = -12\):

\(x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}\)

\(x = \frac{4 \pm \sqrt{16 + 48}}{2}\)

\(x = \frac{4 \pm \sqrt{64}}{2}\)

\(x = \frac{4 \pm 8}{2}\)

\(x = 6\) or \(x = -2\)

For \(x = 6\), the third term is \(x^2 + 12 = 6^2 + 12 = 48\).

For \(x = -2\), the third term is \(x^2 + 12 = (-2)^2 + 12 = 16\).

Let the first term be \(a = -2222\) and the common difference be \(d = 17\). The \(n\)-th term of an arithmetic progression is given by:

\(a_n = a + (n-1) imes d\)

We need to find the smallest \(n\) such that \(a_n > 0\).

\(-2222 + (n-1) imes 17 > 0\)

\((n-1) imes 17 > 2222\)

\(n-1 > \frac{2222}{17}\)

\(n-1 > 130.7\)

Since \(n\) must be an integer, \(n = 132\).

Now, calculate the 132nd term:

\(a_{132} = -2222 + 131 imes 17\)

\(a_{132} = -2222 + 2227\)

\(a_{132} = 5\)

Thus, the first positive term is 5.

Let the first term be \(a = 56\) and the common difference be \(d = 53 - 56 = -3\).

The last term is given as \(-22\). Using the formula for the \(n\)-th term of an arithmetic progression, \(a_n = a + (n-1)d\), we have:

\(-22 = 56 + (n-1)(-3)\)

\(-22 = 56 - 3(n-1)\)

\(-22 = 56 - 3n + 3\)

\(-22 = 59 - 3n\)

\(3n = 59 + 22\)

\(3n = 81\)

\(n = 27\)

The sum of the first \(n\) terms of an arithmetic progression is given by:

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

where \(l\) is the last term.

\(S_{27} = \frac{27}{2} (56 + (-22))\)

\(S_{27} = \frac{27}{2} (34)\)

\(S_{27} = 27 \times 17\)

\(S_{27} = 459\)

Given the arithmetic progression with terms \(a, 2a, a^2\), the common difference \(d\) is:

\(d = 2a - a = a\)

\(d = a^2 - 2a\)

Equating the two expressions for \(d\):

\(a = a^2 - 2a\)

\(a^2 - 3a = 0\)

\(a(a - 3) = 0\)

Since \(a\) is positive, \(a = 3\).

Thus, \(d = 3\).

The sum of the first 50 terms \(S_{50}\) is given by:

\(S_{50} = \frac{50}{2} (2a + 49d)\)

Substitute \(a = 3\) and \(d = 3\):

\(S_{50} = 25 (2 \times 3 + 49 \times 3)\)

\(S_{50} = 25 (6 + 147)\)

\(S_{50} = 25 \times 153\)

\(S_{50} = 3825\)

Let the angle of the smallest sector be \(a\) and the common difference be \(d\). The angles of the sectors are \(a, a+d, a+2d, a+3d, a+4d\).

Given that the largest angle is 4 times the smallest angle:

\(a + 4d = 4a\)

\(3a = 4d\)

The sum of the angles in the circle is 360°:

\(360 = S_5 = \frac{5}{2}(2a + 4d)\)

Substitute \(4d = 3a\) into the equation:

\(360 = \frac{5}{2}(2a + 3a) = 12.5a\)

\(a = 28.8^{\circ}\)

The largest angle is \(a + 4d = 4a = 115.2^{\circ}\).

Given the sum of the first \(n\) terms of an arithmetic progression: \(S_n = 32n - n^2\).

To find the first term \(a\), set \(n = 1\):

\(S_1 = 32(1) - 1^2 = 32 - 1 = 31\).

Thus, the first term \(a = 31\).

To find the common difference \(d\), set \(n = 2\):

\(S_2 = 32(2) - 2^2 = 64 - 4 = 60\).

The second term is given by \(S_2 - S_1 = 60 - 31 = 29\).

The common difference \(d = 29 - 31 = -2\).

Therefore, the first term is 31 and the common difference is -2.

The formula for the nth term of an arithmetic progression is given by:

\(a_n = a + (n-1)d\)

where \(a\) is the first term and \(d\) is the common difference.

Given: \(a = 7\), \(a_n = 84\), and \(a_{3n} = 245\).

For the nth term:

\(7 + (n-1)d = 84\)

\((n-1)d = 77\)

For the (3n)th term:

\(7 + (3n-1)d = 245\)

\((3n-1)d = 238\)

Now, divide the two equations:

\(\frac{n-1}{3n-1} = \frac{77}{238}\)

Simplify \(\frac{77}{238}\) to \(\frac{11}{34}\):

\(\frac{n-1}{3n-1} = \frac{11}{34}\)

Cross-multiply:

\(34(n-1) = 11(3n-1)\)

\(34n - 34 = 33n - 11\)

\(n = 23\)

The sum of the first \(n\) terms of an arithmetic progression is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

For the first 200 terms:

\(S_{200} = \frac{200}{2} (2a + 199d) = 100(2a + 199d)\)

For the first 100 terms:

\(S_{100} = \frac{100}{2} (2a + 99d) = 50(2a + 99d)\)

Given that \(S_{200} = 4S_{100}\):

\(100(2a + 199d) = 4 \times 50(2a + 99d)\)

\(100(2a + 199d) = 200(2a + 99d)\)

\(2a + 199d = 2(2a + 99d)\)

\(2a + 199d = 4a + 198d\)

\(199d - 198d = 4a - 2a\)

\(d = 2a\)

For the 100th term:

The 100th term \(T_{100}\) is given by:

\(T_{100} = a + 99d\)

Substituting \(d = 2a\):

\(T_{100} = a + 99(2a)\)

\(T_{100} = a + 198a\)

\(T_{100} = 199a\)

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

The fifth term is given by:

\(a + 4d = 197\)

The sum of the first ten terms is given by:

\(\frac{10}{2} [2a + 9d] = 2040\)

Simplifying the sum equation:

\(5(2a + 9d) = 2040\)

\(2a + 9d = 408\)

We now have two equations:

1. \(a + 4d = 197\)

2. \(2a + 9d = 408\)

Multiply the first equation by 2:

\(2a + 8d = 394\)

Subtract this from the second equation:

\((2a + 9d) - (2a + 8d) = 408 - 394\)

\(d = 14\)

Thus, the common difference is 14.

(i) The time taken for each mile forms an arithmetic sequence where the first term is 5 minutes (or 300 seconds) and the common difference is 12 seconds. The nth term of an arithmetic sequence is given by:

\(a_n = a + (n-1) imes d\)

where \(a = 300\) and \(d = 12\). We need to find \(n\) such that \(a_n = 540\) (since 9 minutes is 540 seconds):

\(540 = 300 + (n-1) imes 12\)

\(240 = (n-1) imes 12\)

\(n-1 = 20\)

\(n = 21\)

(ii) The total time for 26 miles is the sum of the arithmetic sequence with 26 terms. The sum \(S_n\) of an arithmetic sequence is given by:

\(S_n = rac{n}{2} imes (2a + (n-1) imes d)\)

For \(n = 26\), \(a = 300\), and \(d = 12\):

\(S_{26} = rac{26}{2} imes (2 imes 300 + 25 imes 12)\)

\(S_{26} = 13 imes (600 + 300)\)

\(S_{26} = 13 imes 900 = 11700\) seconds

Convert 11700 seconds to hours and minutes:

\(11700 ext{ seconds} = 3 ext{ hours and } 15 ext{ minutes}\)

The sum of the first ten terms of an arithmetic progression is given by:

\(S_{10} = \frac{10}{2} (2a + 9d) = 400\)

This simplifies to:

\(2a + 9d = 80\)

The sum of the next ten terms (terms 11 to 20) is given by:

\(S_{20} - S_{10} = 1000\)

\(\frac{20}{2} (2a + 19d) - \frac{10}{2} (2a + 9d) = 1000\)

\(2a + 19d = 140\)

We now have two equations:

1. \(2a + 9d = 80\)

2. \(2a + 19d = 140\)

Subtract equation 1 from equation 2:

\((2a + 19d) - (2a + 9d) = 140 - 80\)

\(10d = 60\)

\(d = 6\)

Substitute \(d = 6\) back into equation 1:

\(2a + 9(6) = 80\)

\(2a + 54 = 80\)

\(2a = 26\)

\(a = 13\)

The sum of the first \(n\) terms of an arithmetic progression is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

Given \(S_n = 2n^2 + 8n\), we equate this to the formula:

\(\frac{n}{2} (2a + (n-1)d) = 2n^2 + 8n\)

For \(n = 1\):

\(S_1 = 2(1)^2 + 8(1) = 10\)

Thus, \(a = 10\).

For \(n = 2\):

\(S_2 = 2(2)^2 + 8(2) = 24\)

\(S_2 = a + (a + d) = 24\)

\(10 + (10 + d) = 24\)

\(20 + d = 24\)

\(d = 4\)

Therefore, the first term is 10 and the common difference is 4.

Let the first term be \(a = 12\) and the common difference be \(d\).

The sum of the first four terms is given by:

\(S_4 = \frac{4}{2} (2a + 3d) = 57\)

\(2(24 + 3d) = 57\)

\(48 + 6d = 57\)

\(6d = 9\)

\(d = 1.5\)

The last term \(l = 48\) is given by:

\(l = a + (n-1)d\)

\(48 = 12 + (n-1)1.5\)

\(48 = 12 + 1.5n - 1.5\)

\(48 = 10.5 + 1.5n\)

\(37.5 = 1.5n\)

\(n = 25\)

Let the first term of the arithmetic progression be \(a = 3\) and the common difference be \(d = 2\).

The angles of the sectors form an arithmetic sequence: \(3, 5, 7, \ldots\).

The sum of the angles in a circle is 360°.

The sum of the first \(n\) terms of an arithmetic sequence is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

Substitute \(a = 3\), \(d = 2\), and \(S_n = 360\):

\(\frac{n}{2} (6 + (n-1)2) = 360\)

Simplify:

\(\frac{n}{2} (6 + 2n - 2) = 360\)

\(\frac{n}{2} (2n + 4) = 360\)

\(n(n + 2) = 360\)

\(n^2 + 2n - 360 = 0\)

Solving the quadratic equation \(n^2 + 2n - 360 = 0\) using the quadratic formula:

\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 1\), \(b = 2\), \(c = -360\).

\(n = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-360)}}{2 \times 1}\)

\(n = \frac{-2 \pm \sqrt{4 + 1440}}{2}\)

\(n = \frac{-2 \pm \sqrt{1444}}{2}\)

\(n = \frac{-2 \pm 38}{2}\)

\(n = 18 \text{ or } n = -20\)

Since \(n\) must be positive, \(n = 18\).

(a) The sum of the first \(n\) terms \(S_n\) of an arithmetic progression is given by:

\(S_n = \frac{n}{2} [2a + (n-1)d]\)

Substituting the given values, \(a = 4\) and \(S_n = 5863\):

\(\frac{n}{2} [8 + (n-1)d] = 5863\)

Multiply both sides by 2:

\(n[8 + (n-1)d] = 11726\)

Divide by \(n\):

\(8 + (n-1)d = \frac{11726}{n}\)

Rearrange to find \((n-1)d\):

\((n-1)d = \frac{11726}{n} - 8\)

(b) The \(n\)th term \(u_n\) is given by:

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

Given \(u_n = 139\):

\(4 + (n-1)d = 139\)

Substitute \((n-1)d = \frac{11726}{n} - 8\):

\(\frac{11726}{n} - 8 = 135\)

\(\frac{11726}{n} = 143\)

\(n = \frac{11726}{143} = 82\)

Substitute \(n = 82\) back into \((n-1)d = \frac{11726}{n} - 8\):

\(81d = \frac{11726}{82} - 8\)

\(81d = 143 - 8\)

\(81d = 135\)

\(d = \frac{135}{81} = \frac{5}{3}\)

The first term of the arithmetic progression is given as 61, so \(a = 61\).

The second term is 57, so the common difference \(d\) is \(57 - 61 = -4\).

The formula for the sum of the first \(n\) terms of an arithmetic progression is:

\(S_n = \frac{n}{2} [2a + (n-1)d]\)

Given that the sum \(S_n = n\), we have:

\(n = \frac{n}{2} [2 \times 61 + (n-1)(-4)]\)

Simplifying inside the brackets:

\(n = \frac{n}{2} [122 + (n-1)(-4)]\)

\(n = \frac{n}{2} [122 - 4n + 4]\)

\(n = \frac{n}{2} [126 - 4n]\)

Multiply both sides by 2 to eliminate the fraction:

\(2n = n(126 - 4n)\)

\(2n = 126n - 4n^2\)

Rearrange to form a quadratic equation:

\(4n^2 - 124n = 0\)

Factor out \(n\):

\(n(4n - 124) = 0\)

So, \(n = 0\) or \(4n - 124 = 0\).

Since \(n\) must be positive, solve \(4n - 124 = 0\):

\(4n = 124\)

\(n = 31\)

Thus, the value of the positive integer \(n\) is 31.

Given the sum of the first n terms of an arithmetic progression is \(S_n = n^2 + 8n\).

For the first term \(S_1\):

\(S_1 = 1^2 + 8 \times 1 = 9\)

Thus, the first term \(a = 9\).

For the second term \(S_2\):

\(S_2 = 2^2 + 8 \times 2 = 20\)

We know \(S_2 = a + (a + d)\), so:

\(20 = 9 + (9 + d)\)

\(20 = 18 + d\)

\(d = 2\)

Therefore, the first term is 9 and the common difference is 2.

(i) The sum of an arithmetic progression is given by the formula:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

Given \(S_n = 525\), \(n = 25\), and \(a = -15\), we have:

\(525 = \frac{25}{2} (2(-15) + 24d)\)

\(525 = \frac{25}{2} (-30 + 24d)\)

\(525 = 25(-15 + 12d)\)

\(21 = -15 + 12d\)

\(36 = 12d\)

\(d = 3\)

(ii) The last term \(l\) in the progression is given by:

\(l = a + (n-1)d\)

\(l = -15 + 24 \times 3\)

\(l = -15 + 72\)

\(l = 57\)

(iii) The positive terms start from 3 and go up to 57. The first positive term is 3, which is the second term in the sequence:

\(a + d = 3\)

\(-15 + 3 = 3\)

The number of positive terms is 19 (from 3 to 57). The sum of these terms is:

\(S_{19} = \frac{19}{2} (3 + 57)\)

\(S_{19} = \frac{19}{2} \times 60\)

\(S_{19} = 570\)

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

The sixth term is given by \(a + 5d = 23\).

The sum of the first ten terms is given by \(\frac{10}{2} (2a + 9d) = 200\), simplifying to \(5(2a + 9d) = 200\).

Solving the equations:

1. \(a + 5d = 23\)

2. \(5(2a + 9d) = 200\) simplifies to \(2a + 9d = 40\).

From equation 1, \(a = 23 - 5d\).

Substitute \(a = 23 - 5d\) into equation 2:

\(2(23 - 5d) + 9d = 40\)

\(46 - 10d + 9d = 40\)

\(46 - d = 40\)

\(d = 6\)

Substitute \(d = 6\) back into \(a = 23 - 5d\):

\(a = 23 - 5(6)\)

\(a = 23 - 30\)

\(a = -7\)

The seventh term is \(a + 6d = -7 + 6(6) = -7 + 36 = 29\).

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

The general term of an arithmetic progression is given by \(a + (n-1)d\).

The eighth term is \(a + 7d\) and the third term is \(a + 2d\).

According to the problem, \(a + 7d = 3(a + 2d)\).

Expanding the right side, we have \(a + 7d = 3a + 6d\).

Rearranging gives \(2a = d\).

The sum of the first \(n\) terms of an arithmetic progression is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\).

The sum of the first eight terms is \(S_8 = \frac{8}{2} (2a + 7d) = 4(2a + 7d)\).

The sum of the first four terms is \(S_4 = \frac{4}{2} (2a + 3d) = 2(2a + 3d)\).

Substituting \(d = 2a\) into these expressions:

\(S_8 = 4(2a + 7(2a)) = 4(2a + 14a) = 4(16a) = 64a\).

\(S_4 = 2(2a + 3(2a)) = 2(2a + 6a) = 2(8a) = 16a\).

Thus, \(S_8 = 4 \times S_4\).

Let the angle of the smallest sector be \(a\) and the common difference of the arithmetic progression be \(d\). The angles of the sectors are \(a, a+d, a+2d, a+3d, a+4d, a+5d\).

Given that the largest angle is 4 times the smallest angle, we have:

\(a + 5d = 4a\)

\(5d = 3a\)

The sum of the angles in the circle is \(360^{\circ}\), so:

\(6a + 15d = 360\)

Substitute \(d = \frac{3a}{5}\) into the equation:

\(6a + 15\left(\frac{3a}{5}\right) = 360\)

\(6a + 9a = 360\)

\(15a = 360\)

\(a = 24^{\circ}\)

The arc length of the smallest sector is:

\(\frac{24}{360} \times 2\pi \times 5 = \frac{2\pi}{15} \times 5 = \frac{2\pi}{3}\)

The perimeter of the smallest sector is the sum of the arc length and the two radii:

\(\frac{2\pi}{3} + 2 \times 5 = \frac{2\pi}{3} + 10\)

Approximating \(\pi \approx 3.1416\), the perimeter is:

\(\frac{2 \times 3.1416}{3} + 10 \approx 12.1 \text{ cm}\)

(i) The third term is given by \(a + 2d = 90\) and the fifth term by \(a + 4d = 80\). Solving these equations simultaneously:

Subtract the first from the second: \((a + 4d) - (a + 2d) = 80 - 90\)

\(2d = -10\)

\(d = -5\)

Substitute \(d = -5\) into \(a + 2d = 90\):

\(a + 2(-5) = 90\)

\(a - 10 = 90\)

\(a = 100\)

(ii) The sum of the first \(m\) terms is equal to the sum of the first \((m + 1)\) terms:

\(S_m = S_{m+1}\)

\(\frac{m}{2}(2a + (m-1)d) = \frac{m+1}{2}(2a + md)\)

Since \(S_m = S_{m+1}\), the difference is zero:

\(a + md = 0\)

\(100 + m(-5) = 0\)

\(100 - 5m = 0\)

\(5m = 100\)

\(m = 20\)

(iii) The sum of the first \(n\) terms is zero:

\(S_n = 0\)

\(\frac{n}{2}(2a + (n-1)d) = 0\)

\(n(200 + (n-1)(-5)) = 0\)

\(n(200 - 5n + 5) = 0\)

\(n(205 - 5n) = 0\)

\(205 - 5n = 0\)

\(5n = 205\)

\(n = 41\)

The first term of the arithmetic progression is 161, and the second term is 154. The common difference, d, is given by:

\(d = 154 - 161 = -7\)

The sum of the first m terms of an arithmetic progression is given by the formula:

\(S_m = \frac{m}{2} (2a + (m-1)d)\)

Substituting the given values, we have:

\(S_m = \frac{m}{2} (2 \times 161 + (m-1)(-7)) = 0\)

Simplifying inside the brackets:

\(2 \times 161 = 322\)

Thus, the equation becomes:

\(\frac{m}{2} (322 + (m-1)(-7)) = 0\)

Solving for m:

\(322 + (m-1)(-7) = 0\)

\(322 - 7m + 7 = 0\)

\(329 - 7m = 0\)

\(7m = 329\)

\(m = \frac{329}{7} = 47\)

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

The fifth term of the arithmetic progression is given by:

\(a + 4d = 18\)

The sum of the first 5 terms is given by:

\(\frac{5}{2} (2a + 4d) = 75\)

Solving the first equation for \(a\):

\(a = 18 - 4d\)

Substitute \(a = 18 - 4d\) into the sum equation:

\(\frac{5}{2} (2(18 - 4d) + 4d) = 75\)

\(\frac{5}{2} (36 - 8d + 4d) = 75\)

\(\frac{5}{2} (36 - 4d) = 75\)

\(5(36 - 4d) = 150\)

\(180 - 20d = 150\)

\(30 = 20d\)

\(d = \frac{3}{2}\)

Substitute \(d = \frac{3}{2}\) back into \(a = 18 - 4d\):

\(a = 18 - 4 \times \frac{3}{2}\)

\(a = 18 - 6\)

\(a = 12\)

Thus, the first term is 12, and the common difference is \(\frac{3}{2}\).

The multiples of 5 between 100 and 300 form an arithmetic sequence where the first term is 100 and the common difference is 5.

The sequence is: 100, 105, 110, ..., 300.

To find the number of terms, use the formula for the nth term of an arithmetic sequence:

\(a_n = a + (n-1) imes d\)

where \(a = 100\), \(d = 5\), and \(a_n = 300\).

Solving for \(n\):

\(300 = 100 + (n-1) imes 5\)

\(300 = 100 + 5n - 5\)

\(300 = 95 + 5n\)

\(205 = 5n\)

\(n = 41\)

Now, use the formula for the sum of an arithmetic sequence:

\(S_n = \frac{n}{2} imes (a + a_n)\)

Substitute the known values:

\(S_{41} = \frac{41}{2} imes (100 + 300)\)

\(S_{41} = \frac{41}{2} imes 400\)

\(S_{41} = 41 imes 200\)

\(S_{41} = 8200\)

Therefore, the sum of all the multiples of 5 between 100 and 300 inclusive is 8200.

(a) For an arithmetic progression, the difference between consecutive terms is constant. Therefore, we have:

\(6k - k = k + 6 - 6k\)

\(5k = k + 6 - 6k\)

\(5k = k + 6 - 6k\)

\(5k + 6k = k + 6\)

\(11k = k + 6\)

\(10k = 6\)

\(k = \frac{6}{10} = 0.6\)

(b) The common difference \(d\) is \(6k - k = 5k\). Substituting \(k = 0.6\), we get \(d = 3\).

The sum of the first 30 terms \(S_{30}\) is given by:

\(S_{30} = \frac{30}{2} (2 \times 0.6 + 29 \times 3)\)

\(S_{30} = 15 (1.2 + 87)\)

\(S_{30} = 15 \times 88.2\)

\(S_{30} = 1323\)

(i) The formula for the nth term of an arithmetic progression is given by:

\(a_n = a + (n-1)d\)

For the ninth term, \(a + 8d = 22\).

The sum of the first 4 terms is given by:

\(S_4 = \frac{4}{2}(2a + 3d) = 49\)

\(2a + 3d = 24.5\)

We have the system of equations:

1. \(a + 8d = 22\)

2. \(2a + 3d = 24.5\)

Solving these equations simultaneously, we find:

\(d = 1.5\)

\(a = 10\)

(ii) The nth term is given by:

\(a + (n-1)d = 46\)

Substituting \(a = 10\) and \(d = 1.5\):

\(10 + (n-1) \times 1.5 = 46\)

\(1.5(n-1) = 36\)

\(n-1 = 24\)

\(n = 25\)

The first term of the arithmetic progression is given as \(a = 5\), and the second term is \(9\), so the common difference \(d = 9 - 5 = 4\).

The general formula for the \(n\)-th term of an arithmetic progression is \(a_n = a + (n-1)d\).

We need to find the smallest \(n\) such that \(a_n > 200\).

Set \(a_n = 200\):

\(200 = 5 + (n-1) imes 4\)

\(200 = 5 + 4n - 4\)

\(200 = 4n + 1\)

\(199 = 4n\)

\(n = \frac{199}{4} = 49.75\)

Since \(n\) must be an integer, we take \(n = 50\) as the smallest integer greater than 49.75.

Now, calculate the sum of the first 50 terms using the sum formula for an arithmetic progression:

\(S_n = \frac{n}{2} (a + a_n)\)

\(S_{50} = \frac{50}{2} (5 + 201)\)

\(S_{50} = 25 imes 206\)

\(S_{50} = 5150\)

The first term of the arithmetic progression is given as \(a = 6\).

The fifth term is given by the formula \(a + 4d = 12\).

Substituting \(a = 6\), we have:

\(6 + 4d = 12\)

\(4d = 6\)

\(d = 1.5\)

The sum of the first \(n\) terms of an arithmetic progression is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

We know \(S_n = 90\), \(a = 6\), and \(d = 1.5\), so:

\(\frac{n}{2} (2 \times 6 + (n-1) \times 1.5) = 90\)

\(\frac{n}{2} (12 + 1.5n - 1.5) = 90\)

\(\frac{n}{2} (10.5 + 1.5n) = 90\)

\(n(10.5 + 1.5n) = 180\)

\(1.5n^2 + 10.5n - 180 = 0\)

Divide the entire equation by 1.5:

\(n^2 + 7n - 120 = 0\)

Solving this quadratic equation using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 7\), \(c = -120\):

\(n = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-120)}}{2 \times 1}\)

\(n = \frac{-7 \pm \sqrt{49 + 480}}{2}\)

\(n = \frac{-7 \pm \sqrt{529}}{2}\)

\(n = \frac{-7 \pm 23}{2}\)

\(n = 8\) or \(n = -15\)

Since \(n\) must be positive, \(n = 8\).

To find the sum of all integers between 100 and 400 that are divisible by 7, we first identify the smallest and largest integers in this range that are divisible by 7.

The smallest integer greater than 100 that is divisible by 7 is 105, since 105 is the first multiple of 7 after 100.

The largest integer less than 400 that is divisible by 7 is 399, since 399 is the last multiple of 7 before 400.

These integers form an arithmetic sequence where the first term is 105, the last term is 399, and the common difference is 7.

The number of terms in this sequence, denoted as \(n\), can be found using the formula for the \(n\)-th term of an arithmetic sequence:

\(l = a + (n-1) imes d\)

where \(l = 399\), \(a = 105\), and \(d = 7\).

Solving for \(n\):

\(399 = 105 + (n-1) imes 7\)

\(399 - 105 = (n-1) imes 7\)

\(294 = (n-1) imes 7\)

\(n-1 = \frac{294}{7}\)

\(n-1 = 42\)

\(n = 43\)

The sum of an arithmetic sequence is given by:

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

Substituting the known values:

\(S_{43} = \frac{43}{2} imes (105 + 399)\)

\(S_{43} = \frac{43}{2} imes 504\)

\(S_{43} = 43 imes 252\)

\(S_{43} = 10,836\)

Therefore, the sum of all integers between 100 and 400 that are divisible by 7 is 10,836.

The arithmetic progression (AP) is given by the sequence: 180, 175, 170, \ldots, 25.

The first term \(a = 180\) and the common difference \(d = 175 - 180 = -5\).

The last term \(l = 25\).

To find the number of terms \(n\), use the formula for the last term of an AP: \(l = a + (n-1) \cdot d\).

Substitute the known values: \(25 = 180 + (n-1) \cdot (-5)\).

Simplify: \(25 = 180 - 5(n-1)\).

\(25 = 180 - 5n + 5\).

\(25 = 185 - 5n\).

\(5n = 185 - 25\).

\(5n = 160\).

\(n = \frac{160}{5} = 32\).

Now, use the formula for the sum of an AP: \(S_n = \frac{n}{2} (a + l)\).

Substitute the known values: \(S_{32} = \frac{32}{2} (180 + 25)\).

\(S_{32} = 16 \cdot 205\).

\(S_{32} = 3280\).

Let the first term of the arithmetic progression be \(a = 60\) and the common difference be \(d\). The total number of payments is \(n = 48\), and the total amount repaid is \(S_n = 3726\).

The formula for the sum of an arithmetic progression is:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

Substitute the given values:

\(3726 = \frac{48}{2} (2 \times 60 + 47d)\)

\(3726 = 24 (120 + 47d)\)

\(3726 = 2880 + 1128d\)

\(846 = 1128d\)

\(d = \frac{846}{1128} = 0.75\)

The third payment is given by the formula for the nth term of an arithmetic progression:

\(a_3 = a + 2d\)

\(a_3 = 60 + 2 \times 0.75\)

\(a_3 = 60 + 1.5\)

\(a_3 = 61.50\)

Let the first term be \(a = -10\) and the common difference be \(d\).

The 15th term is given by \(a + 14d = 11\).

Substituting \(a = -10\), we have:

\(-10 + 14d = 11\)

\(14d = 21\)

\(d = \frac{3}{2}\)

The last term is given by \(a + (n-1)d = 41\).

Substituting \(a = -10\) and \(d = \frac{3}{2}\), we have:

\(-10 + (n-1) \times \frac{3}{2} = 41\)

\((n-1) \times \frac{3}{2} = 51\)

\(n-1 = 34\)

\(n = 35\)

The sum of the arithmetic progression is given by:

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

\(S_{35} = \frac{35}{2} (-10 + 41)\)

\(S_{35} = \frac{35}{2} \times 31\)

\(S_{35} = 542.5\)

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

The formula for the \(n\)-th term of an arithmetic progression is \(a_n = a + (n-1)d\).

Given the 13th term is 12, we have:

\(a + 12d = 12\) (Equation 1)

The formula for the sum of the first \(n\) terms is \(S_n = \frac{n}{2} (2a + (n-1)d)\).

Given the sum of the first 30 terms is -15, we have:

\(\frac{30}{2} (2a + 29d) = -15\)

\(15(2a + 29d) = -15\)

\(2a + 29d = -1\) (Equation 2)

Solving Equations 1 and 2 simultaneously:

From Equation 1: \(a = 12 - 12d\)

Substitute into Equation 2:

\(2(12 - 12d) + 29d = -1\)

\(24 - 24d + 29d = -1\)

\(5d = -25\)

\(d = -5\)

Substitute \(d = -5\) back into Equation 1:

\(a + 12(-5) = 12\)

\(a - 60 = 12\)

\(a = 72\)

Now, find the sum of the first 50 terms:

\(S_{50} = \frac{50}{2} (2a + 49d)\)

\(S_{50} = 25 (2(72) + 49(-5))\)

\(S_{50} = 25 (144 - 245)\)

\(S_{50} = 25 (-101)\)

\(S_{50} = -2525\)

(a) The \(n\)th term of an arithmetic progression is given by \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.

Here, \(a = 84\) and \(d = -3\). We need \(a_n < 0\).

\(84 + (n-1)(-3) < 0\)

\(84 - 3(n-1) < 0\)

\(84 - 3n + 3 < 0\)

\(87 < 3n\)

\(n > 29\)

The smallest integer \(n\) is 30.

(b) The sum of the first \(m\) terms of an arithmetic progression is given by \(S_m = \frac{m}{2} [2a + (m-1)d]\).

We are given \(S_{2k} = S_k\).

\(\frac{2k}{2} [2 \times 84 + (2k-1)(-3)] = \frac{k}{2} [2 \times 84 + (k-1)(-3)]\)

\(k [168 + (2k-1)(-3)] = \frac{k}{2} [168 + (k-1)(-3)]\)

\(k [168 - 6k + 3] = \frac{k}{2} [168 - 3k + 3]\)

\(k [171 - 6k] = \frac{k}{2} [171 - 3k]\)

\(2k [171 - 6k] = k [171 - 3k]\)

\(342k - 12k^2 = 171k - 3k^2\)

\(342k - 171k = 12k^2 - 3k^2\)

\(171k = 9k^2\)

\(19k = k^2\)

\(k = 19\)

For the arithmetic progression P, the 5th term is given by:

\(a + 4d.\)

For the arithmetic progression Q, the 12th term is given by:

\(2(a + 1) + 11(d + 1).\)

Using the given ratio:

\(\frac{a + 4d}{2(a + 1) + 11(d + 1)} = \frac{1}{3}.\)

Cross-multiplying gives:

\(3(a + 4d) = 2(a + 1) + 11(d + 1).\)

Simplifying, we get:

\(a + d = 13.\)

For the sum of the first 5 terms of P:

\(\frac{5}{2}(2a + 4d).\)

For the sum of the first 5 terms of Q:

\(\frac{5}{2}(4(a + 1) + 4(d + 1)).\)

Using the given ratio:

\(\frac{\frac{5}{2}(2a + 4d)}{\frac{5}{2}(4(a + 1) + 4(d + 1))} = \frac{2}{3}.\)

Cross-multiplying gives:

\(3(2a + 4d) = 2(4(a + 1) + 4(d + 1)).\)

Simplifying, we get:

\(-a + 2d = 8.\)

Solving the equations \(a + d = 13\) and \(-a + 2d = 8\) simultaneously, we find:

\(d = 7\) and \(a = 6\).

The sum of the first n terms of an arithmetic progression is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

For the first 20 terms:

\(\frac{20}{2} (2a + 19d) = 405\)

\(10(2a + 19d) = 405\)

For the first 40 terms:

\(\frac{40}{2} (2a + 39d) = 1410\)

\(20(2a + 39d) = 1410\)

We have the equations:

1. \(10(2a + 19d) = 405\)

2. \(20(2a + 39d) = 1410\)

Solving these simultaneously:

From equation 1: \(2a + 19d = 40.5\)

From equation 2: \(2a + 39d = 70.5\)

Subtract equation 1 from equation 2:

\(20d = 30\)

\(d = 1.5\)

Substitute \(d = 1.5\) into equation 1:

\(2a + 19(1.5) = 40.5\)

\(2a + 28.5 = 40.5\)

\(2a = 12\)

\(a = 6\)

Now, find the 60th term \(T_{60}\):

\(T_{60} = a + 59d\)

\(T_{60} = 6 + 59(1.5)\)

\(T_{60} = 6 + 88.5\)

\(T_{60} = 94.5\)