Exam practice

The terms between the first and last terms of an arithmetic sequence. For example, in 2, 5, 8, 11 the numbers 5 and 8 are the two arithmetic means between 2 and 11.

m − a = b − m, so $m = \frac{a + b}{2}$.

$\frac{12 + 18}{2} =$ 15.

$\frac{6 + 10}{2} =$ 8.

$\frac{-2 + 2}{2} =$ 0.

$\frac{-4 + 8}{2} =$ 2.

$\frac{(a + b) + (a - b)}{2} = \frac{2a}{2} =$ a.

n = 6: 20 = 5 + 5d, d = 3. 8, 11, 14, 17.

n = 8: 14 = 70 + 7d, d = −8. 62, 54, 46, 38, 30, 22.

n = 8: −9 = 5 + 7d, d = −2. 3, 1, −1, −3, −5, −7.

$x = \frac{5 + 9}{2} =$ 7.

Equal differences: (x + 5) − (x + 1) = (3x + 1) − (x + 5), so 4 = 2x − 4 and x = 4. (The sequence is 5, 9, 13.)

3x − (x + 2) = (4x + 1) − 3x, so 2x − 2 = x + 1 and x = 3. (The sequence is 5, 9, 13.)

a = 7, d = 4: $S_{20} = 10[14 + 19 \times 4] = 10 \times 90 =$ 900.

a = 4, d = −5: $S_7 = \frac{7}{2}[8 + 6 \times (-5)] = \frac{7}{2} \times (-22) =$ −77.

$a = \frac12$, d = 1: $S_{16} = 8[1 + 15] =$ 128.

65 = 5 + (n − 1)5, so n = 13. $S_{13} = \frac{13}{2}(5 + 65) =$ 455.

a = −64, d = 16: 32 = −64 + (n − 1)16, so n = 7. $S_7 = \frac{7}{2}(-64 + 32) =$ −112.

1 + 3 + 5 + ⋯: $S_{10} = 5[2 + 9 \times 2] =$ 100. (The first n odd numbers always add up to $n^2$.)

$S_{100} = \frac{100}{2}(1 + 100) =$ 5,050.

From 50 to 100 there are 100 − 50 + 1 = 51 numbers. $S = \frac{51}{2}(50 + 100) =$ 3,825.

$204 = \frac{n}{2}(1 + 50)$, so n = 8. 50 = 1 + 7d: d = 7.

$440 = \frac{n}{2}(60 + 20)$, so n = 11. 20 = 60 + 10d: d = −4.

$a + l = 17 - \frac{99}{8} = \frac{37}{8}$, so $\frac{407}{16} = \frac{n}{2} \times \frac{37}{8} = \frac{37n}{16}$ and n = 11.
$-\frac{99}{8} = 17 + 10d$, so $10d = -\frac{235}{8}$ and $d = -\frac{47}{16}$.

325 = 5[2a + 9 × (−3)], so 2a − 27 = 65: a = 46.

$108 = \frac{9}{2}[2a + 72]$, so 2a + 72 = 24: a = −24.

155 = 5[2a + 27], so 2a + 27 = 31: a = 2.

$374 = \frac{n}{2}[8 + 6(n - 1)] = n(3n + 1)$, so $3n^2 + n - 374 = 0$, (n − 11)(3n + 34) = 0: 11 terms.

$540 = \frac{n}{2}[72 + 9(n - 1)]$, so $9n^2 + 63n - 1080 = 0$, $n^2 + 7n - 120 = 0$, (n − 8)(n + 15) = 0: 8 terms.

a + 2d = −15 and a + 7d = 10: 5d = 25, d = 5, a = −25.
$S_{16} = 8[-50 + 15 \times 5] = 8 \times 25 =$ 200.

a + 4d = 10 and a + 10d = 22: 6d = 12, d = 2, a = 2.
$S_{20} = 10[4 + 19 \times 2] =$ 420.

3(2a + 5d) = 75, so 2a + 5d = 25; 6(2a + 11d) = 390, so 2a + 11d = 65.
Subtracting: 6d = 40, $d = \frac{20}{3}$; then $2a = 25 - \frac{100}{3} = -\frac{25}{3}$.
$S_{20} = 10\left(2a + 19d\right) = 10\left(-\frac{25}{3} + \frac{380}{3}\right) = \frac{3550}{3} =$ $1183\tfrac13$.

$\frac{7}{2}(2a + 6d) = 21$, so a + 3d = 3; 6(2a + 11d) = 126, so 2a + 11d = 21.
From the first, 2a + 6d = 6; subtracting, 5d = 15, d = 3, a = −6.
The series is −6 − 3 + 0 + 3 + 6 + ⋯

$S_{10} = 5[2000 + 9 \times 100] = 5 \times 2900 = 14500$.
No: he makes 14,500, 500 short of 15,000.

First year a = 60000 × 12 = 720000, d = 24000.
$10224000 = \frac{n}{2}[1440000 + 24000(n - 1)]$, so $20448000 = 24000n^2 + 1416000n$, and dividing by 24,000: $n^2 + 59n - 852 = 0$.
(n − 12)(n + 71) = 0: 12 years.

The terms between the first and last terms of a geometric sequence. For example, in 2, 6, 18, 54 the numbers 6 and 18 are the two geometric means between 2 and 54.

$\frac{m}{a} = \frac{b}{m}$, so $m^2 = ab$ and $m = \sqrt{ab}$.

$\sqrt{3 \times 27} = \sqrt{81} =$ 9.

$m^2 = (-4)(-64) = 256$, so m = ±16.
Both make a geometric sequence: −4, −16, −64 (ratio 4) and −4, 16, −64 (ratio −4).

$\sqrt{\frac15 \times 125} = \sqrt{25} =$ 5.

$\sqrt{7 \times 343} = \sqrt{2401} =$ 49.

n = 6: $192 = 6r^5$, $r^5 = 32$, r = 2. 12, 24, 48, 96.

n = 5: $405 = 5r^4$, $r^4 = 81$, r = 3. 15, 45, 135.

n = 5: $\frac49 = \frac94 r^4$, so $r^4 = \frac{16}{81}$ and $r = \frac23$.
$\frac32$, 1, $\frac23$. (With $r = -\frac23$: $-\frac32$, 1, $-\frac23$.)

$x^2 = 4 \times 9 = 36$: x = ±6 (4, 6, 9 with ratio $\frac32$, or 4, −6, 9 with ratio $-\frac32$).

$r = \frac84 = 2$, so $x = \frac42 =$ 2.

r = 5, so x + 1 = 125: x = 124.

$S_6 = \frac{2(2^6 - 1)}{2 - 1} = 2 \times 63 =$ 126.

$a = \frac19$, r = 3: $S_5 = \frac{\frac19(243 - 1)}{2} = \frac{242}{18} =$ $\frac{121}{9}$.

$a = -\frac14$, r = −2: $S_6 = \frac{a(1 - r^6)}{1 - r} = \frac{-\frac14(1 - 64)}{3} = \frac{-\frac14 \times (-63)}{3} =$ $\frac{21}{4}$.

a = 16, $r = \frac12$, $l = \frac{1}{16}$: $S = \frac{a - lr}{1 - r} = \frac{16 - \frac{1}{32}}{\frac12} = 32 - \frac{1}{16} =$ $31\tfrac{15}{16}$.

a = 1, $r = \frac13$, $l = \frac{1}{729}$: $S = \frac{1 - \frac{1}{2187}}{\frac23} = \frac{2186}{2187} \times \frac32 = \frac{1093}{729} =$ $1\tfrac{364}{729}$.

$728 = \frac{486r - 2}{r - 1}$, so 728r − 728 = 486r − 2, 242r = 726: r = 3.

1820r − 1820 = 1215r − 5, 605r = 1815: r = 3.

1533r − 1533 = 768r − 3, 765r = 1530: r = 2.

ar = 4 and $ar^6 = 128$: dividing, $r^5 = 32$, r = 2, a = 2.
$S_{10} = 2(2^{10} - 1) =$ 2,046.

ar = 3 and $ar^4 = 81$: dividing, $r^3 = 27$, r = 3, a = 1.
$S_7 = \frac{3^7 - 1}{2} = \frac{2186}{2} =$ 1,093.

$r = \frac{48}{32} = \frac32$: $665 = \frac{32\left[\left(\frac32\right)^n - 1\right]}{\frac12} = 64\left[\left(\tfrac32\right)^n - 1\right]$.
So $\left(\frac32\right)^n = \frac{665}{64} + 1 = \frac{729}{64} = \left(\frac32\right)^6$: 6 terms.

a = 6, r = −2: $-2046 = \frac{6[1 - (-2)^n]}{3} = 2[1 - (-2)^n]$.
So $1 - (-2)^n = -1023$, $(-2)^n = 1024 = (-2)^{10}$: 10 terms.

$43680 = \frac{a(3^6 - 1)}{2} = 364a$, so a = 120.
Last: $ar^5 = 120 \times 243 = 29160$. Difference: 29160 − 120 = Rs 29,040.