Arithmetic Progressions — Class 10 Maths PYQs

(d) $15$

(c) $6$

(b) $3$

(C) $0$

(C) $-2a - 4d$

(b) $\sqrt{96}$

(a) $3$

(d) $\sqrt{112}$

(b) $3$

(B) $17$

(c) $7$

(A) $12$

(A) 4

(B) $-2$

(B) $-2$

(D) $5$

(C) $\sqrt{162}$

(D) $\sqrt{507}$

(A) $\sqrt{2}$

(C) $d - c$

(c) $p+1$

(D) $3$

(C) $3$

(A) $59$

(C) $5$ (1 Mark)

(B) $5$

(C) Assertion (A) is true, but Reason (R) is false.

(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

Here $a = 32$, $l = -10$ and $d = -2$
$\therefore 32 + (n - 1)(-2) = -10$ (1/2 Mark)
$\Rightarrow n = 22$ (1/2 Mark)
S$_{22} = \frac{22}{2} \times [32 + (-10)]$ (1/2 Mark)
$= 242$ (1/2 Mark)

$a_1 = 3 (1) - 2 = 1$ (1/2 Mark)
and $a_{28} = 3 (28) - 2 = 82$ (1/2 Mark)
S$_{28} = \frac{28}{2} \times (1+82)$ (1/2 Mark)
$= 1162$ (1/2 Mark)

$a + 2d = 16 \dots(i)$ ($\frac{1}{2}$ mark)
$a + 6d = 12 + a + 4d \dots(ii)$ ($\frac{1}{2}$ mark)
Solving (i) and (ii) to get $d = 6, a = 4$ ($\frac{1}{2}$ mark)
$\therefore \text{A.P. is } 4, 10, 16, \dots$ ($\frac{1}{2}$ mark)
$S_{29} = \frac{29}{2} [8 + 28 \times 6] = 2552$ ($\frac{1}{2} + \frac{1}{2}$ marks)

Given $\frac{a + 10d}{a + 16d} = \frac{3}{4}$
$\Rightarrow 4a + 40d = 3a + 48d$
$\Rightarrow a = 8d$ (i)∴ $\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d} = \frac{3}{7}$ using(i)
$a_5: a_{21} = 3:7$
$\frac{S_5}{S_{21}} = \frac{\frac{5}{2} (2a + 4d)}{\frac{21}{2} (2a + 20d)} = \frac{5 \times 20d}{21 \times 36d} = \frac{25}{189}$
Therefore, $S_5:S_{21}=25:189$

$-4+(-1)+2 + 5 + ..... + x = 437$
Here a = $-4$, d = $3$
$-4 + (n - 1)3 = x \Rightarrow n = \frac{x + 7}{3}$
$S_n = 437$
$\Rightarrow \frac{n}{2}(a+x) = 437$
$\Rightarrow \frac{(\frac{x+7}{3})}{2}(-4+x) = 437$
$\frac{(x+7)(x-4)}{6} = 437$
$x^2 + 3x - 28 = 437 \times 6 = 2622$
$x^2 + 3x - 2650 = 0$
$(x + 53)(x - 50) = 0$
$x \ne -53, x = 50$

Let first term = $a$, common difference = $$\begin{aligned}& d = 2 \\ & a_n = a + (n - 1)2 = 4 \\ & text{ATQ, } a + 2n = 6 \\ & a = 6-2n \\ & text{ATQ, } S_n = \frac{n}{2}[2a + (n - 1)2] = -14 \\ & n[a + n - 1] = -14 \\ & n[6 - 2n + n - 1] = -14 \\ & n^2 - 5n - 14 = 0 \\ & Rightarrow n = 7 \\ & text{and } a = -8 \\ & S_{20} = \frac{20}{2}[2\times (-8) + 19 \times 2] \\ & = 220\end{aligned}$$

(B) $10$

Both (B) $3$ and (D) $-3$ are correct options

(b) $4$-digit numbers, divisible by $7$ are $1001, 1008, 1015, \text{......}, 9996$ (½ Mark)
It is an AP with $a = 1001, d = 7, a_n = 9996$ (1 Mark)
$9996 = 1001 + (n - 1)(7)$ (½ Mark)
$\Rightarrow (n-1)7 = 8995 \Rightarrow n-1 = 1285 \Rightarrow n = 1286$ (½ Mark)

Interest at the end of $1^{st}$ year $= \frac{2000 \times 7 \times 1}{100} = \text{Rs}140$
Interest at the end of $2^{nd}$ year $= \frac{2000 \times 7 \times 2}{100} = \text{Rs}280$
Interest at the end of $3^{rd}$ year $= \frac{2000 \times 7 \times 3}{100} = \text{Rs}420$
$140, 280, 420, \dots$
Yes, Interests form an AP with first term = $140$ and common difference = $140$
Interest at the end of $27^{th}$ year $= 140 + 26 \times 140$
$= \text{Rs}3780$

Here $a = 50$ and $d = 20$
(i) Number of trees planted in $10^{th}$ row $= a_{10} = 50 + 9 \times 20 = 230$
(ii) $a_8 - a_5 = 3 \times 20 = 60$
(iii) (a) Let $S_n = 3200$
$\Rightarrow \frac{n}{2}[2 \times 50 + (n-1) \times 20] = 3200$
$\Rightarrow n^2 + 4n - 320 = 0$
$\Rightarrow (n + 20)(n - 16) = 0$
$n \neq -20$
$\therefore n = 16$
Hence, required number of rows are $16$
OR
(iii) (b) Required number of trees $= S_n - S_{11}$
$= 3200 - \frac{11}{2}[2 \times 50 + 10 \times 20]$
$= 3200 - 1550$
$= 1650$
Hence, number of trees left are $1650$

(i) $$\begin{aligned}& (a + 8d) - (a + 5d) = 330 \Rightarrow d = 110 \\ & a+5\times110 = 800 \Rightarrow a = 250 \\ & (ii) d=110 \\ & (iii) (a)\end{aligned}$$a₈ = 250 + 7 × 110
= 1020
OR
(iii) (b) $$\begin{aligned}& S_6 = \frac{6}{2} [2 \times 250 + 5 \times 110] \\ & = 3150\end{aligned}$$

A.P: $300, 350, 400 \dots$ (i) $a_4 = 450, a_5 = 500, a_6 = 550$. (ii) $a_8 = 300 + 7 \times 50 = 650$ m. (iii)(a) $S_{10} = \frac{10}{2}(2 \times 300 + 9 \times 50) = 5250$ m. (iii)(b) $S_6 = \frac{6}{2}(2 \times 300 + 5 \times 50) = 2250$ m.

The ages of the participants form the following AP
$8, 8\frac{1}{3}, 8\frac{2}{3}, 9, ...$
where first term $= 8$ and common difference $= \frac{1}{3}$
Let the number of participants be $n$
$S_n = \frac{n}{2}[2 \times 8 + (n-1)\frac{1}{3}] = 168$
$n^2 + 47n - 1008 = 0$
$\Rightarrow n = 16$
$\therefore$ the age of the eldest participant $= 8 + 15 \times \frac{1}{3} = 13$ years

(i) Number of dots formed in each square are $4, 8, 12, ...$
$12 - 8 = 8 - 4 = 4$
$\therefore$ Numbers of dots form an A.P. (I) ($\frac{1}{2}$ Mark)
Here, $a = 4, d = 4$ (II) ($\frac{1}{2}$ Mark)
(ii) $a_n = 4 + (n - 1) 4 = 4n$ (I) (1 Mark)
(iii) (a) Here, $a = 4, d = 4, S_n = 220$
$\therefore 220 = \frac{n}{2} (2 \times 4 + (n - 1)4)$ (I) ($\frac{1}{2}$ Mark)
$\Rightarrow (n + 1) n = 110$
$\Rightarrow n^2 + n - 110 = 0$
$\Rightarrow (n + 11) (n - 10) = 0$
$\Rightarrow n = -11, 10$
$n \neq -11$
$\Rightarrow n = 10$ (II) ($\frac{1}{2}$ Mark)
$\therefore$ The number of squares formed $= 10$
OR
(b) Here $S_n = 100, a = 4, d = 4$
$\therefore 100 = \frac{n}{2} (2 \times 4 + (n - 1)4)$ (I) ($\frac{1}{2}$ Mark)
$\Rightarrow 100 = \frac{4n}{2} (n + 1)$
$\Rightarrow n (n + 1) = 50$
$\Rightarrow n^2 + n - 50 = 0$
$\Rightarrow n = \frac{-1 \pm \sqrt{1+200}}{2} = \frac{-1 \pm \sqrt{201}}{2}$ (II) ($\frac{1}{2}$ Mark)
$\therefore n$ is not a natural number.
$\therefore$ It is not possible to draw complete squares using $100$ dots. (III) (1 Mark)

Total amount = ₹$5,90,000$
First instalment = ₹$5,000$ and Increase in the instalment = ₹$500$
(a) ₹$5,000$ ; ₹$5,500$ ; ₹$6,000$ (1 Mark)
(b) $11^{th}$ instalment ($a_{11}$) = $5000 + 10 \times 500$ (1 Mark)
$= 10000$
Amount paid in $11^{th}$ instalment is ₹$10,000$
(c) Let '$n$' be the number of instalments to clear the loan.
$590000 = \frac{n}{2} [2 \times 5000 + (n - 1)500]$ (1 Mark)
$\Rightarrow n^2 + 19n - 2360 = 0$ ($\frac{1}{2}$ Mark)
$\Rightarrow (n + 59)(n - 40) = 0$ ($\frac{1}{2}$ Mark)
Thus, $n = -59, 40$
rejecting $n = -59$,
$\therefore n = 40$ ($\frac{1}{2}$ Mark)
Thus, the loan will be cleared in $40$ instalments. ($\frac{1}{2}$ Mark)
OR
(c) Amount to be paid upto $31^{st}$ instalment is $S_{31}$ ($\frac{1}{2}$ Mark)
$S_{31} = \frac{31}{2} [2 \times 5000 + (31 - 1)500]$ (1 Mark)
$= \frac{31}{2} \times 25000$
$= 387500$ ($\frac{1}{2}$ Mark)
Amount still to be paid = $5,90,000 - 3,87,500$ ($\frac{1}{2}$ Mark)
$= \text{Rs}2,02,500$ ($\frac{1}{2}$ Mark)

(i) The number of members at the level eleven $= a_{11}$ (1 Mark)
$= 1$
(ii) The number of members at the third level from the top $= a_9$ (1 Mark)
$= 9$
(iii) (a) The total number of members who formed the human pyramid
$S_{11} = \frac{11}{2} [41+1]$ (1/2 Mark)
$= 231$ (1/2 Mark)
OR
(iii) (b) Let $n^{th}$ level be the required level.
$a_n = 5 a_{10}$ (1/2 Mark)
$41 + (n - 1) (– 4) = 5 [41 + 9 \times (-4)]$ (1 Mark)
$n = 5$ (1/2 Mark)
At $5^{th}$ level the number of members is $5$ times the number of members at level ten

Here $a = 22, d = – 1$ $S_n = 250$
$\therefore 250 = \frac{n}{2} [44 + (n – 1) (-1)]$
$\Rightarrow n^2 - 45n + 500 = 0$
$\Rightarrow (n-25) (n – 20) = 0$
$n\neq 25 \therefore n = 20$
logs in top row = $a_{20} = 22 + 19 (- 1) = 3$

Let his starting salary is $a$ and annual increment be $d$.
A.T.Q
$a+3d=15000$ ---------(i)
$a+9d=18000$ ----------(ii)
On solving equations (i) & (ii)
$a = 13500$ & $d = 500$
Starting salary = ₹13,500
Annual increment = ₹500

(D) $10$

(B) $\frac{19}{12}$