NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.3#

$a = 2$, $d = 5$, $n = 10$.

$a = -37$, $d = 4$, $n = 12$.

$a = 0.6$, $d = 1.1$, $n = 100$.

$a = \dfrac{1}{15}$, $d = \dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5-4}{60} = \dfrac{1}{60}$, $n = 11$.

$a = 7$, $d = 3.5$, $l = 84$.

First find $n$: $84 = 7 + (n-1)(3.5) \implies 77 = 3.5(n-1) \implies n-1 = 22 \implies n = 23$

$a = 34$, $d = -2$, $l = 10$.

$10 = 34 + (n-1)(-2) \implies -24 = -2(n-1) \implies n = 13$

$a = -5$, $d = -3$, $l = -230$.

$-230 = -5 + (n-1)(-3) \implies -225 = -3(n-1) \implies n = 76$

$50 = 5 + (n-1)(3) \implies 45 = 3(n-1) \implies n = 16$

$35 = 7 + 12d \implies 12d = 28 \implies d = \dfrac{7}{3}$

$37 = a + 11(3) \implies a = 37 - 33 = 4$

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

$S_{10} = \dfrac{10}{2}[2a + 9d] = 5(2a+9d) = 125 \implies 2a + 9d = 25 \quad \cdots (2)$

From (1): $a = 15 - 2d$. Substituting in (2):

$a = 15 + 2 = 17$

$a_{10} = 17 + 9(-1) = \mathbf{8}$

$S_9 = \dfrac{9}{2}[2a + 8(5)] = \dfrac{9}{2}[2a+40] = 75$

$2a + 40 = \dfrac{150}{9} = \dfrac{50}{3}$

let use: $S_9 = 9a + 36d = 75 \implies 9a + 180 = 75 \implies 9a = -105 \implies a = -\dfrac{35}{3}$

$a_9 = a + 8d = -\dfrac{35}{3} + 40 = \dfrac{-35+120}{3} = \dfrac{85}{3}$

$90 = \dfrac{n}{2}[4 + (n-1)(8)] = \dfrac{n}{2}[8n - 4] = n(4n-2)$

$4n^2 - 2n - 90 = 0 \implies 2n^2 - n - 45 = 0$

$D = 1 + 360 = 361$, $\sqrt{D} = 19$

$n = \dfrac{1+19}{4} = 5$ (taking positive root)

$a_5 = 2 + 4(8) = 2 + 32 = \mathbf{34}$

$S_n = \dfrac{n}{2}(a + a_n) \implies 210 = \dfrac{n}{2}(8+62) = 35n \implies n = 6$

$62 = 8 + 5d \implies d = \dfrac{54}{5} = \mathbf{10.8}$

$a_n = a + (n-1)(2) = 4 \implies a = 4 - 2(n-1) = 6 - 2n \quad \cdots (1)$

$S_n = \dfrac{n}{2}(a + 4) = -14 \implies n(a+4) = -28 \quad \cdots (2)$

From (1): $a + 4 = 10 - 2n$. Substituting in (2): $n(10-2n) = -28 \implies 10n - 2n^2 = -28 \implies n^2 - 5n - 14 = 0$

$(n-7)(n+2) = 0 \implies n = 7$ (taking positive value)

$a = 6 - 14 = \mathbf{-8}$

$192 = \dfrac{8}{2}[6 + 7d] = 4(6+7d) = 24 + 28d$

$28d = 168 \implies d = \mathbf{6}$

$S = \dfrac{n}{2}(a+l) \implies 144 = \dfrac{9}{2}(a+28) \implies a + 28 = 32 \implies a = \mathbf{4}$

$a = 9$, $d = 8$, $S_n = 636$.

$D = 25 + 10176 = 10201$, $\sqrt{D} = 101$

∴ 12 terms must be taken.

$a = 5$, $l = 45$, $S_n = 400$.

$S_n = \dfrac{n}{2}(a+l) \implies 400 = \dfrac{n}{2}(50) = 25n \implies n = 16$

$l = a + (n-1)d \implies 45 = 5 + 15d \implies d = \dfrac{40}{15} = \dfrac{8}{3}$

∴ $n = \mathbf{16}$ and $d = \dfrac{8}{3}$

$a = 17$, $l = 350$, $d = 9$.

$350 = 17 + (n-1)(9) \implies 333 = 9(n-1) \implies n = 38$

$a_{22} = a + 21(7) = 149 \implies a = 149 - 147 = 2$

$d = 18 - 14 = 4$, $a_2 = 14 \implies a = 10$.

$S_7 = \dfrac{7}{2}(2a+6d) = 7(a+3d) = 49 \implies a + 3d = 7 \quad \cdots (1)$

$S_{17} = \dfrac{17}{2}(2a+16d) = 17(a+8d) = 289 \implies a + 8d = 17 \quad \cdots (2)$

Subtract (1) from (2): $5d = 10 \implies d = 2$, $a = 1$.

$a_n = 4n + 3$

$a_1 = 7$, $a_2 = 11$, $a_3 = 15$

$a_{n+1} - a_n = [4(n+1)+3] - [4n+3] = 4$ (constant)

∴ This is an AP with $a = 7$ and $d = 4$.

$a_n = 9 - 5n$

$a_1 = 4$, $a_2 = -1$, $a_3 = -6$

$a_{n+1} - a_n = [9-5(n+1)] - [9-5n] = -5$ (constant)

∴ This is an AP with $a = 4$ and $d = -5$.

$S_n = 4n - n^2$

$S_1 = 4(1) - 1 = \mathbf{3}$ (= first term $a_1 = 3$)

$S_2 = 4(2) - 4 = 4$ (sum of first 2 terms)

$a_2 = S_2 - S_1 = 4 - 3 = \mathbf{1}$

$a_3 = S_3 - S_2 = (12-9) - 4 = 3 - 4 = \mathbf{-1}$

$a_{10} = S_{10} - S_9 = (40-100) - (36-81) = -60 - (-45) = \mathbf{-15}$

$a_n = S_n - S_{n-1} = (4n-n^2) - [4(n-1)-(n-1)^2]$ $= 4n - n^2 - 4n + 4 + n^2 - 2n + 1 = \mathbf{5 - 2n}$

The AP: $6, 12, 18, \ldots$ ($a = 6$, $d = 6$, $n = 40$)

The AP: $8, 16, 24, \ldots$ ($a = 8$, $d = 8$, $n = 15$)

Odd numbers: $1, 3, 5, \ldots, 49$

$a = 1$, $d = 2$, $l = 49$. $n = 25$.

$a = 200$, $d = 50$, $n = 30$.

Let the first (highest) prize = $₹a$. $d = -20$, $n = 7$, $S_7 = 700$.

The prizes are: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.

Trees planted by one section of class $n$ = $n$ trees.

Since there are 3 sections per class: Total trees by class $n$ = $3n$.

Total trees by all 12 classes:

Length of $n$th semicircle $= \pi r_n$ where $r_n = 0.5n$ cm.

Total length $= \pi \sum_{n=1}^{13} 0.5n = \dfrac{\pi}{2} \cdot \dfrac{13 \times 14}{2} = \dfrac{\pi}{2} \times 91 = \dfrac{91\pi}{2}$

Using $\pi = \dfrac{22}{7}$:

$a = 20$, $d = -1$, $S_n = 200$.

$D = 1681 - 1600 = 81$, $\sqrt{D} = 9$

$n = 25$ or $n = 16$.

If $n = 25$: top row $= 20 + 24(-1) = -4$ (impossible).

So $n = 16$: Top row $= 20 + 15(-1) = \mathbf{5}$ logs.

∴ The logs are stacked in 16 rows with 5 logs in the top row.

Distance for 1st potato (to and fro) $= 2 \times 5 = 10$ m

Distance for 2nd potato $= 2 \times (5+3) = 16$ m

Distance for 3rd potato $= 2 \times (5+6) = 22$ m

$\vdots$

Distance for $n$th potato $= 2(5 + 3(n-1)) = 2(3n+2) = 6n+4$

The distances form an AP: $10, 16, 22, \ldots$ with $a = 10$, $d = 6$, $n = 10$.