$t_{10} = 5(10) – 3 = 50 – 3 = 47$

$t_{15} = 5(15) – 3 = 75 – 3 = 72$

$5n – 3 = 607 \Rightarrow 5n = 610 \Rightarrow n = 122$

Apply rule repeatedly:

Sequence: $-5, -2, 1, 4, 7, \ldots$ This is an AP with $a=-5$, $d=3$.
Explicit rule: $t_n = -5+(n-1)\times3 = 3n-8$

Is 52 a term? $3n-8=52 \Rightarrow 3n=60 \Rightarrow n=20$.

Each term = sum of the three previous terms:

$t_n = a + (n-1)d = 3 + (n-1) \times 5$

$t_{10} = 3 + 9 \times 5 = 3 + 45 = 48$
$t_{26} = 3 + 25 \times 5 = 3 + 125 = 128$

$a = 21$, $d = 18-21 = -3$
$t_n = 21 + (n-1)(-3) = 21 – 3n + 3 = 24 – 3n$

$a = 11$, $d = 8-11 = -3$

Explicit rule:
$t_n = 11 + (n-1)(-3) = 11 – 3n + 3 = 14 – 3n$

Recursive rule:
$t_1 = 11,\quad t_n = t_{n-1} – 3 \text{ for } n \geq 2$

$t_3 = a + 2d = 12 \quad \cdots(1)$
$t_{50} = a + 49d = 106 \quad \cdots(2)$

Subtract (1) from (2):
$47d = 94 \Rightarrow d = 2$

From (1): $a = 12 – 4 = 8$

$t_{29} = a + 28d = 8 + 28 \times 2 = 8 + 56 = 64$

2-digit multiples of 3: $12, 15, 18, \ldots, 99$
$a = 12$, $d = 3$, last term $= 99$

Find $n$: $12 + (n-1)\times3 = 99 \Rightarrow (n-1) = 29 \Rightarrow n = 30$

Sum $= \dfrac{n}{2}(a + l) = \dfrac{30}{2}(12 + 99) = 15 \times 111 = 1665$

$a = 5,00,000$, $d = 20,000$
$t_n = 5,00,000 + (n-1) \times 20,000 = 7,00,000$
$(n-1) \times 20,000 = 2,00,000$
$n – 1 = 10 \Rightarrow n = 11$

Total $= 1 + 2 + 3 + \ldots + 25 = S_{25} = \dfrac{25 \times 26}{2} = \dfrac{650}{2} = 325$

Given: $r = 2$, $t_8 = 192$
$t_8 = a \cdot r^7 = a \cdot 2^7 = 128a = 192$
$\therefore a = \dfrac{192}{128} = \dfrac{3}{2}$

$t_{12} = a \cdot r^{11} = \dfrac{3}{2} \times 2^{11} = \dfrac{3}{2} \times 2048 = 3 \times 1024 = 3072$

$a = 5$, $r = 25/5 = 5$

$t_n = 5 \times 5^{n-1} = 5^n$

$t_{10} = 5^{10} = 9,\!765,\!625$

Build terms step by step:

$a = 2$, $r = 6/2 = 3$
$t_n = 2 \times 3^{n-1} = 4374$
$3^{n-1} = 2187 = 3^7 \Rightarrow n-1 = 7 \Rightarrow n = 8$

$a = 2$, $r = 2\sqrt{2}/2 = \sqrt{2} = 2^{1/2}$

$t_n = 2 \times (\sqrt{2})^{n-1} = 2^1 \times 2^{(n-1)/2} = 2^{1+(n-1)/2}$

Set $t_n = 128 = 2^7$:
$1 + \dfrac{n-1}{2} = 7 \Rightarrow \dfrac{n-1}{2} = 6 \Rightarrow n-1 = 12 \Rightarrow n = 13$

$a + 10d = 38 \quad \cdots(1)$
$a + 15d = 73 \quad \cdots(2)$

Subtract: $5d = 35 \Rightarrow d = 7$
From (1): $a = 38 – 70 = -32$

$t_{31} = a + 30d = -32 + 30 \times 7 = -32 + 210 = 178$

$t_7 – t_5 = 2d = 12 \Rightarrow d = 6$
$t_3 = a + 2d = a + 12 = 16 \Rightarrow a = 4$

AP: $4, 10, 16, 22, 28, \ldots$

Smallest 3-digit multiple of 7: $7 \times 15 = 105$
Largest 3-digit multiple of 7: $7 \times 142 = 994$

AP: $105, 112, 119, \ldots, 994$ with $a=105$, $d=7$

$105 + (n-1) \times 7 = 994$
$(n-1) \times 7 = 889 \Rightarrow n-1 = 127 \Rightarrow n = 128$

Smallest multiple of 4 $> 10$: $12$
Largest multiple of 4 $< 250$: $248$

AP: $12, 16, 20, \ldots, 248$ with $a=12$, $d=4$

$12 + (n-1) \times 4 = 248$
$(n-1) = 59 \Rightarrow n = 60$

$t_5 = 4t_3 \Rightarrow ar^4 = 4ar^2 \Rightarrow r^2 = 4 \Rightarrow r = 2$ or $r = -2$

Case 1: $r = 2$
$a + ar = a(1+r) = a(3) = -4 \Rightarrow a = -\dfrac{4}{3}$
GP: $-\dfrac{4}{3},\ -\dfrac{8}{3},\ -\dfrac{16}{3},\ -\dfrac{32}{3},\ \ldots$

Case 2: $r = -2$
$a(1-2) = -a = -4 \Rightarrow a = 4$
GP: $4,\ -8,\ 16,\ -32,\ 64,\ \ldots$

Sum of $k$ consecutive integers starting from $m$:
$km + \dfrac{k(k-1)}{2} = 100 \Rightarrow km = 100 – \dfrac{k(k-1)}{2}$

Test values of $k$:

GP: $a = 30$, $r = 2$, $t_n = 30 \times 2^n$ (at end of $n$th hour)

End of 2nd hour: $30 \times 2^2 = 30 \times 4 = 120$
End of 4th hour: $30 \times 2^4 = 30 \times 16 = 480$
End of $n$th hour: $30 \times 2^n$

$(a+3d)+(a+7d) = 24 \Rightarrow 2a+10d = 24 \Rightarrow a+5d = 12 \quad\cdots(1)$
$(a+5d)+(a+9d) = 44 \Rightarrow 2a+14d = 44 \Rightarrow a+7d = 22 \quad\cdots(2)$

Subtract (1) from (2): $2d = 10 \Rightarrow d = 5$
From (1): $a = 12-25 = -13$

First three terms: $-13,\ -8,\ -3$

$S_n = \dfrac{n(n+1)}{2} > 1000 \Rightarrow n(n+1) > 2000$

Try $n = 44$: $44 \times 45 = 1980 < 2000$ ✗
Try $n = 45$: $45 \times 46 = 2070 > 2000$ ✓

$S_{44} = 1980/2 = 990 \leq 1000$
$S_{45} = 2070/2 = 1035 > 1000$ ✓

$a = 2$, $r = 4$; $t_n = 2 \times 4^{n-1}$

$2 \times 4^{n-1} = 131072 = 2^{17}$
$2^1 \times 2^{2(n-1)} = 2^{17}$
$1 + 2(n-1) = 17 \Rightarrow 2n-1 = 17 \Rightarrow n = 9$

Let terms be $\dfrac{a}{r},\ a,\ ar$.

Product: $\dfrac{a}{r} \cdot a \cdot ar = a^3 = -1 \Rightarrow a = -1$

Sum: $\dfrac{-1}{r} + (-1) + (-r) = \dfrac{13}{12}$
$-\dfrac{1}{r} – r = \dfrac{13}{12} + 1 = \dfrac{25}{12}$
$\dfrac{1}{r} + r = -\dfrac{25}{12}$

$12r^2 + 25r + 12 = 0$
$r = \dfrac{-25 \pm \sqrt{625-576}}{24} = \dfrac{-25 \pm 7}{24}$

$r = -\dfrac{3}{4}$ or $r = -\dfrac{4}{3}$

For $r = -\dfrac{3}{4}$: Terms = $\dfrac{4}{3},\ -1,\ \dfrac{3}{4}$

Let GP have first term $A$ and common ratio $R$.
$x = AR^3,\quad y = AR^9,\quad z = AR^{15}$

Check if $y^2 = xz$:
$y^2 = (AR^9)^2 = A^2R^{18}$
$xz = AR^3 \cdot AR^{15} = A^2R^{18}$

$\therefore y^2 = xz$ $\Rightarrow x, y, z$ are in GP. $\quad\blacksquare$

Let terms be $\dfrac{a}{r},\ a,\ ar$.

Sum: $a\!\left(\dfrac{1}{r}+1+r\right) = 26 \quad\cdots(1)$
Sum of squares: $a^2\!\left(\dfrac{1}{r^2}+1+r^2\right) = 364 \quad\cdots(2)$

Let $S = \dfrac{1}{r}+1+r$. Then $\dfrac{1}{r^2}+1+r^2 = S^2 – 2S$

From (2): $a^2(S^2-2S) = 364$; from (1): $aS = 26$, so $a = \dfrac{26}{S}$

$\left(\dfrac{26}{S}\right)^2 S(S-2) = 364 \Rightarrow \dfrac{676(S-2)}{S} = 364$
$676S – 1352 = 364S \Rightarrow 312S = 1352 \Rightarrow S = \dfrac{13}{3}$

$a = \dfrac{26 \times 3}{13} = 6$

$\dfrac{1}{r} + r = \dfrac{10}{3} \Rightarrow 3r^2 – 10r + 3 = 0$
$r = 3$ or $r = \dfrac{1}{3}$

For $r=3$: Terms = $2, 6, 18$

Compute step by step:

Pattern: $1, 2, 4, 8, 16, 32, 64, 128$ — powers of 2!

Simpler recursive formula: $P_n = 2P_{n-1}$ for $n > 2$
(Since $P_n – P_{n-1} = P_{n-1}$, i.e., each term doubles)

Explicit formula: $P_n = 2^{n-1}$ for $n \geq 1$

Compute step by step: