Progressions & Series: CAT previous year questions

Solution

There are 29 odd numbers from 1 to 57, with total $29^2 = 841$.

Let $k$ be the $j$-th odd number ($k = 2j - 1$). Sum below $= (j - 1)^2$; sum above $= 841 - j^2$.

$(j - 1)^2 = 841 - j^2 \Rightarrow j^2 - j - 420 = 0 \Rightarrow j = 21$.

$k = 2(21) - 1 = 41$.

Solution

Terms $a, ar, ar^2$: $a(1 + r + r^2) = 52$.

Pairwise products: $a^2 r(1 + r + r^2) = 624 \Rightarrow ar \times 52 = 624 \Rightarrow ar = 12$.

$\dfrac{12}{r} + 12 + 12r = 52 \Rightarrow 3r^2 - 10r + 3 = 0 \Rightarrow r = \dfrac{1}{3}$ (decreasing), $a = 36$.

Sum $= \dfrac{36}{1 - 1/3} = 54$.

Solution

$a_4 + a_7 + a_{10} = 3a_7 = 99 \Rightarrow a_7 = 33$.

$S_{14} = 7(a_1 + a_{14}) = 7(a_7 + a_8) = 497 \Rightarrow a_8 = 38$, so $d = 5$ and $a_1 = 3$.

$S_5 = \dfrac{5}{2}(2 \times 3 + 4 \times 5) = 65$.

Solution

Let the first term be $a$ and the common difference $d$.

$2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d) \Rightarrow 2a + 4d = 0 \Rightarrow a = -2d$.

$x_5 = a + 4d = 2d = -4 \Rightarrow d = -2$, $a = 4$.

$x_{100} = 4 + 99(-2) = -194$.

Solution

The $n$-th term is $\left(\dfrac{1}{25}\right)^n - \left(\dfrac{1}{35}\right)^n$.

Sum $= \dfrac{1/25}{1 - 1/25} - \dfrac{1/35}{1 - 1/35} = \dfrac{1}{24} - \dfrac{1}{34} = \dfrac{10}{816} = \dfrac{5}{408}$.

Solution

Even terms: $t_4 = \dfrac{1}{3}t_2$, $t_6 = \dfrac{3}{5}t_4$, …, so $t_{2k} = -\dfrac{1}{2k - 1}$.

$\dfrac{1}{t_{2k}} = -(2k - 1)$.

Sum for $k = 1$ to $1012$ $= -(1 + 3 + \dots + 2023) = -1012^2 = -1024144$.

Solution

Let $a = \sqrt{x}$, $b = \sqrt{y}$, $c = \sqrt{z}$. Given $\dfrac{2}{b + c} = \dfrac{1}{a + c} + \dfrac{1}{a + b}$.

$2(a + b)(a + c) = (b + c)(2a + b + c)$.

Expanding and cancelling: $2a^2 = b^2 + c^2 \Rightarrow 2x = y + z$.

So $y, x, z$ are in arithmetic progression.

Solution

$N_n = 2N_{n-1} + 3 \Rightarrow N_n + 3 = 2(N_{n-1} + 3)$. With $N_1 + 3 = 5$: $N_n = 5 \times 2^{n-1} - 3$.

Need $5 \times 2^{n-1} > 1000003 \Rightarrow 2^{n-1} > 200000.6$.

$2^{17} = 131072$ ✗, $2^{18} = 262144$ ✓, so $n - 1 = 18$.

Lowest $n = 19$.

Solution

Let the common differences be primes $p$ (for $a$) and $q$ (for $b$). $b_2 = 0 \Rightarrow b_1 = -q$.

$a_5 = b_9$: $a_1 + 4p = 7q$. $a_{19} = b_{19}$: $a_1 + 18p = 17q$.

Subtracting: $14p = 10q \Rightarrow 7p = 5q \Rightarrow p = 5$, $q = 7$.

$a_1 = 49 - 20 = 29$; $a_{11} = 29 + 10 \times 5 = 79$.

Solution

$a_n = 6n + 7$: all terms $\equiv 1 \pmod 6$ (from 13). $b_n = 7n + 8$: all terms $\equiv 1 \pmod 7$ (from 15).

Common terms are $\equiv 1 \pmod{42}$: $43, 85, \dots$

Largest three-digit: $42 \times 23 + 1 = 967$.

Solution

The series is $\sum_{n \geq 0} \left(\dfrac{1}{4}\right)^n \sum_{k=0}^{n} \left(\dfrac{1}{3}\right)^k$.

Swap the order: $\sum_{k \geq 0} \left(\dfrac{1}{3}\right)^k \sum_{n \geq k} \left(\dfrac{1}{4}\right)^n = \sum_{k} \left(\dfrac{1}{3}\right)^k \left(\dfrac{1}{4}\right)^k \cdot \dfrac{4}{3}$.

$= \dfrac{4}{3} \times \dfrac{1}{1 - 1/12} = \dfrac{4}{3} \times \dfrac{12}{11} = \dfrac{16}{11}$.

Solution

$a_n$: 54, 62, …, 846 (all $\equiv 6 \pmod 8$). $b_n$: 102, 106, …, 498 (all $\equiv 2 \pmod 4$).

Every $a_n$ is $\equiv 2 \pmod 4$, so the common terms are the $a_n$ between 102 and 498: 102, 110, …, 494.

That is 50 terms; sum $= \dfrac{50(102 + 494)}{2} = 14900$.

Solution

$t_n = S_n - S_{n-1} = (n + 2n^2) - [(n - 1) + 2(n - 1)^2] = 4n - 1$.

$4n - 1$ divisible by 9: $n = 7$ gives 27 ✓ (smaller $n$ give 3, 7, 11, 15, 19, 23).

Smallest $n = 7$.

Solution

On day $n$ the count is multiplied by $1 + \dfrac{1}{n} = \dfrac{n + 1}{n}$.

After day $m$: $100 \times \dfrac{3}{2} \times \dfrac{4}{3} \times \dots \times \dfrac{m + 1}{m} = 100 \times \dfrac{m + 1}{2}$.

$50(m + 1) = 1000 \Rightarrow m = 19$.

Solution

$A_n = \dfrac{n}{2}[6 + 4(n - 1)] = 2n^2 + n$.

$\sum_{n=1}^{25} A_n = 2 \times \dfrac{25 \times 26 \times 51}{6} + \dfrac{25 \times 26}{2} = 11050 + 325 = 11375$.

$\dfrac{11375}{25} = 455$.

Solution

Terms are $38 + 17k$. First 3-digit term: $k = 4$, giving 106. Last: $38 + 17k \leq 999 \Rightarrow k = 56$, giving 990.

These form an AP, so the average $= \dfrac{106 + 990}{2} = 548$.

Solution

Compute: $x_0 = 1$, $x_1 = 2$, $x_2 = 3$, $x_3 = 2$, $x_4 = 1$, $x_5 = 1$, $x_6 = 2$, …

The sequence repeats with period 5.

$2021 = 5 \times 404 + 1$, so $x_{2021} = x_1 = 2$.

Solution

Group $k$ has $2k - 1$ numbers and ends at $k^2$.

Group 15 runs from $14^2 + 1 = 197$ to $15^2 = 225$: 29 numbers.

Sum $= \dfrac{29(197 + 225)}{2} = 29 \times 211 = 6119$.

Solution

Let the numbers be $y - d, y, y + d$. Then $xyz = y(y^2 - d^2) = 5 \times 3y \Rightarrow y^2 - d^2 = 15$.

$(y - d)(y + d) = 15 = 1 \times 15$ or $3 \times 5$. Since $d > 2$: $y - d = 1$, $y + d = 15$, so $d = 7$.

$z - x = 2d = 14$.

Solution

$(-1)^{n+1}x_n = (n^2 + 2n) - ((n - 1)^2 + 2(n - 1)) = 2n + 1$.

$x_{49} = 99$ (odd index, plus sign). $-x_{50} = 101 \Rightarrow x_{50} = -101$.

$x_{49} + x_{50} = -2$.

Solution

$x_{100} = x_1 + \sum_{n=1}^{99}(n - 1) = -1 + (0 + 1 + \dots + 98)$.

$= -1 + \dfrac{98 \times 99}{2} = -1 + 4851 = 4850$.

Solution

$r^{n - m} = \dfrac{12}{3/4} = 16$, with $n - m$ a positive integer.

Options: $r = 2, n - m = 4$ (sum 6); $r = 4, 2$ (6); $r = 16, 1$ (17); $r = -2, 4$ (2); $r = -4, 2$ ($-2$).

Smallest $r + n - m = -2$.

Solution

$x_{m+1} = x_m - (m + 1)$.

$x_{100} = x_1 - (2 + 3 + \dots + 100) = -1 - 5049 = -5050$.