Chapter 8 - Sequences and Series

Given $S_p = 0$. Sum formula: $\frac{p}{2}[2a + (p-1)d] = 0$.
Since $p \neq 0$, $2a + (p-1)d = 0 \implies d = \frac{-2a}{p-1}$.
The sum of the next $q$ terms is $S_{p+q} - S_p = S_{p+q} - 0 = S_{p+q}$.
$S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]$.
Substitute $d$: $S_{p+q} = \frac{p+q}{2} \left[ 2a + (p+q-1)\left(\frac{-2a}{p-1}\right) \right]$
$= \frac{p+q}{2} \cdot 2a \left[ 1 - \frac{p+q-1}{p-1} \right] = a(p+q) \left[ \frac{p-1 - (p+q-1)}{p-1} \right]$
$= a(p+q) \left[ \frac{-q}{p-1} \right] = -\frac{a(p+q)q}{p-1}$.

Let the saving in the first year be $a$. The savings form an A.P. with $n=20$ and $d=200$.
Total savings $S_{20} = 66000$.
$S_{20} = \frac{20}{2}[2a + (20-1)d] = 10[2a + 19(200)]$.
$66000 = 10[2a + 3800] \implies 6600 = 2a + 3800$.
$2a = 2800 \implies a = 1400$.

This is an A.P. with $a = 5200$ and $d = 320$.
(a) Salary for 10th month: $a_{10} = a + 9d = 5200 + 9(320) = 5200 + 2880 = 8080$.
(b) Total earnings in first year (12 months): $S_{12} = \frac{12}{2}[2a + 11d] = 6[2(5200) + 11(320)] = 6[10400 + 3520] = 6[13920] = 83520$.

Let first term be $A$ and common ratio be $R$.
$AR^{p-1} = q$ and $AR^{q-1} = p$.
Dividing: $\frac{AR^{p-1}}{AR^{q-1}} = \frac{q}{p} \implies R^{p-q} = \frac{q}{p} \implies R = (\frac{q}{p})^{\frac{1}{p-q}}$.
From first eq: $A = \frac{q}{R^{p-1}}$.
$(p+q)^{th}$ term $T_{p+q} = AR^{p+q-1} = \frac{q}{R^{p-1}} R^{p+q-1} = q R^{(p+q-1)-(p-1)} = q R^q$.
Substitute $R$: $T_{p+q} = q \left( (\frac{q}{p})^{\frac{1}{p-q}} \right)^q = q \frac{q^{\frac{q}{p-q}}}{p^{\frac{q}{p-q}}} = \frac{q^{1 + \frac{q}{p-q}}}{p^{\frac{q}{p-q}}} = \frac{q^{\frac{p-q+q}{p-q}}}{p^{\frac{q}{p-q}}} = \frac{q^{\frac{p}{p-q}}}{p^{\frac{q}{p-q}}} = \left( \frac{q^p}{p^q} \right)^{\frac{1}{p-q}}$.

This is an A.P. with $a=5, d=2$. Total frames $S_n = 192$.
$S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[10 + (n-1)2] = \frac{n}{2}[2n + 8] = n(n+4)$.
$n^2 + 4n = 192 \implies n^2 + 4n - 192 = 0$.
$(n+16)(n-12) = 0$.
Since $n$ cannot be negative, $n = 12$.

Sum of angles for $n$-sided polygon is $(n-2) \times 180^\circ$.
For $n=3, 4, 5, \dots$, sums are $180^\circ, 360^\circ, 540^\circ, \dots$.
This is an A.P. with $a=180^\circ$ and $d=180^\circ$.
For a 21-sided polygon ($n=21$), the sum is $(21-2) \times 180^\circ = 19 \times 180^\circ = 3420^\circ$.

Side of 1st triangle = 20 cm. Perimeter $P_1 = 60$ cm.
Side of 2nd triangle (joining midpoints) = $20/2 = 10$ cm. Perimeter $P_2 = 30$ cm.
Side of 3rd triangle = $10/2 = 5$ cm. Perimeter $P_3 = 15$ cm.
Perimeters form a G.P.: $60, 30, 15, \dots$ with $a=60, r=1/2$.
Perimeter of 6th triangle $P_6 = ar^5 = 60(1/2)^5 = \frac{60}{32} = \frac{15}{8} = 1.875$ cm.

Distance for 1st potato: $2 \times 24 = 48$ m.
Distance for 2nd potato: $2 \times (24+4) = 56$ m.
Distance for 3rd potato: $2 \times (24+8) = 64$ m.
This is an A.P. with $a=48, d=8, n=20$.
Total distance $S_{20} = \frac{20}{2}[2(48) + 19(8)] = 10[96 + 152] = 10[248] = 2480$ m.

Let the prizes be $a_1, a_2, \dots, a_{16}$ (from last to first).
This forms an A.P. with $n=16$, total sum $S_{16} = 8000$.
Lowest prize (last place) $a = 275$. Let the highest prize (first place) be $l$.
$S_{16} = \frac{16}{2}(a + l) \implies 8000 = 8(275 + l)$.
$1000 = 275 + l \implies l = 725$.

Rationalize each term: $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k+1}-a_k} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$.
Sum $= \frac{1}{d} [(\sqrt{a_2}-\sqrt{a_1}) + (\sqrt{a_3}-\sqrt{a_2}) + \dots + (\sqrt{a_n}-\sqrt{a_{n-1}})]$.
Terms cancel out (telescoping series), leaving $\frac{1}{d} [\sqrt{a_n} - \sqrt{a_1}]$.
Multiply numerator and denominator by $\sqrt{a_n} + \sqrt{a_1}$:
$= \frac{a_n - a_1}{d(\sqrt{a_n} + \sqrt{a_1})}$.
Since $a_n = a_1 + (n-1)d$, we have $a_n - a_1 = (n-1)d$.
Sum $= \frac{(n-1)d}{d(\sqrt{a_n} + \sqrt{a_1})} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$.

The $k^{th}$ term is $T_k = (2k+1)^3 - (2k)^3$.
$T_k = (8k^3 + 12k^2 + 6k + 1) - 8k^3 = 12k^2 + 6k + 1$.
(i) Sum $S_n = \sum_{k=1}^n (12k^2 + 6k + 1) = 12 \sum k^2 + 6 \sum k + \sum 1$.
$= 12 \frac{n(n+1)(2n+1)}{6} + 6 \frac{n(n+1)}{2} + n$
$= 2n(n+1)(2n+1) + 3n(n+1) + n$
$= n [ 2(2n^2+3n+1) + 3n+3 + 1 ]$
$= n [ 4n^2 + 6n + 2 + 3n + 4 ] = n(4n^2 + 9n + 6) = 4n^3 + 9n^2 + 6n$.
(ii) For $n=10$: $4(1000) + 9(100) + 6(10) = 4000 + 900 + 60 = 4960$.

Given $S_n = 3n^2 + 2n$.
$n^{th}$ term $a_n = S_n - S_{n-1}$.
$a_n = (3n^2 + 2n) - [3(n-1)^2 + 2(n-1)]$
$= 3n^2 + 2n - [3(n^2 - 2n + 1) + 2n - 2]$
$= 3n^2 + 2n - [3n^2 - 6n + 3 + 2n - 2]$
$= 3n^2 + 2n - [3n^2 - 4n + 1]$
$= 6n - 1$.
Therefore, the $r^{th}$ term is $6r - 1$.