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) $\sqrt{112}$

(D) $\sqrt{507}$

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

(a) $\sqrt{2}$ (1 Mark)

(D) $1^2, 3^2, 5^2, 7^2, ...$

(A) $6$ (1 Mark)

(A) $\sqrt{2}$

(C) $d - c$

(c) $5$

(c) $4$

(c) $p+1$

(c) $2$

(B) $a-2, b-2$ and $c-2$ are also in A.P.

(D) $2p$, $2q$ and $2r$ are also in A.P.

$x-3, y-3, z-3$

(B) $5$

Answer (C) $-12$

Answer (D) $\frac{34}{9}$

(D) $3$

(b) Let the three consecutive terms in A.P. be $a - d, a, a + d$
∴ $(a - d) + a + (a + d) = 21$
$a = 7$ (1/2 Mark)
and $a (a^2 - d^2) = 231 \Rightarrow d^2 = 16$
$d = \pm 4$ (1/2 Mark)
When $d = 4$ the terms are $3, 7, 11$
When $d = -4$ the terms are $11, 7, 3$ (1 Mark)

(a) $40$

(c) $-32$

(d) $-64$

(C) $31$

(C) $0$

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(B) $16^{th}$

(C) $37$

(B) $10$

(C) $16^{th}$

(C) $3$

(C) $137$

(B) $43$

(D) $3$

(A) $\frac{11}{4}$

(C) $115$

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

(c) $p + 9q$ (1 Mark)

(A) $59$

(B) $n-\frac{4}{3}$ (1 Mark)

(A) $56$ (1 Mark)

(C) $5$ (1 Mark)

(B) $4$

(B) $\frac{5}{3}$

(c) $-66\sqrt{2}$

(b) $8$

(A) $-75$

(B) $-40\sqrt{3}$

(C) $\frac{-39}{2}$

(D) $-\frac{43}{3}$

(C) $115$

(B) $\sqrt{200}$

(B) $3n - \frac{9}{2}$

Answer (B) $\frac{5n-6}{3}$

Answer (C) $56$

$a = -14, a_5 = 2 \Rightarrow a+4d=2$
$-14 + 4d = 2 \Rightarrow d = 4$
$a_n = 62 \Rightarrow a + (n-1)d = 62$
$-14 + (n - 1)4 = 62 \Rightarrow n = 20$

$65, 61, 57, 53, \dots$
$a = 65, d = -4$
Let $a_n$ be the first negative term
$a_n < 0 \Rightarrow a+(n-1)d < 0$
$65 + (n - 1) (-4) <0 \Rightarrow 69 – 4n < 0$
$n > \frac{69}{4}$
$\therefore$ Least positive integral value of $n$ which satisfies $n > \frac{69}{4}$ is $18$
$\therefore 1^{st}$ negative term of the AP = $18$

$a_3 + a_7 = 40$
$(a + 2d) + (a + 6d) = 40$
$\Rightarrow 2a + 8d = 40$ or $a + 4d = 20$ --- (1)
$a_6 + a_{14} = 70$
$(a + 5d) + (a + 13d) = 70$
$\Rightarrow 2a +18d = 70$ or $a + 9d = 35$ --- (2)
Solving (1) and (2), we get
$a = 8$ and $d = 3$
$S_{10} = \frac{10}{2} \times [2 (8) + 9 (3)]$
$= 215$

$a_n = 9 + (n-1) 6 = 243$
$\implies n = 40$
$a_{20} = 9 + 19 \times 6 = 123$
$a_{21} = 9 + 20 \times 6 = 129$
$\therefore$ The two middle terms are $123$ and $129$

$a + 3d = 13$ (i) (1/2 Mark)
$a + 9d = 25$ (ii) (1/2 Mark)
solving (i) & (ii) to get $d = 2, a = 7$ (1 Mark)
therefore $a_{24} = 53$ (1 Mark)

Multiples of $3$ from $232$ to $540$ are $234, 237, 240, ......, 540$ (1 Mark)
This forms an AP with $a=234, d=3$ (1 Mark)
$540 = 234 + (n - 1)3$
$\Rightarrow n = 103$ (1 Mark)

(C) $20100$

(A) $7$

(B) $5$

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

(B) $4$

(B) $9$

(C) $0$

(C) $a_{10} = -a$

(A) $2$

(B) $5$

(a) $S_n = \frac{n}{2}[2a + (n - 1)d] = 90$ (1 Mark)
$\frac{n}{2}[4 + (n - 1)8] = 90 \Rightarrow n[2 + 4n - 4] = 90 \Rightarrow n[4n - 2] = 90 \Rightarrow 4n^2 - 2n - 90 = 0 \Rightarrow 2n^2 - n - 45 = 0$ (1 Mark)
$\Rightarrow (2n+9)(n-5) = 0 \Rightarrow n = 5, -\frac{9}{2}$ (rejected) (½ Mark)
$\Rightarrow n = 5$ (½ Mark)

(a) Here $a = 3, d = 2, S_n = 80$ (1 Mark)
$\frac{n}{2}[6 + (n - 1)2] = 80$ (1/2 Mark)
$n^2 + 2n - 80 = 0$
$n = 8, -10$ (rejected)
$n = 8$ (1/2 Mark)

Let $a$ be the first term and $d$ be the common difference.
$\frac{7}{2}(2a + 6d) = 49$
$a+3d=7 \dots (i)$
$\frac{17}{2}(2a + 16d) = 289$
$a + 8d = 17 \dots (ii)$
solving (i) and (ii)
$d=2 \& a=1$
$S_{20} = \frac{20}{2} [2(1)+19(2)]$
$= 400$

$45, 39, 33, .......$
a = $45$, d = $- 6$
Sn = $180$
$180 = \frac{n}{2} [2 \times 45 + (n - 1) (-6)]$
$180 = \frac{n}{2} [90-6n + 6]$
$\Rightarrow 360 = 96n - 6n^2$
$\Rightarrow 6n^2 - 96n + 360 = 0$
$\Rightarrow n^2 - 16n + 60 = 0 \Rightarrow (n - 10) (n - 6) = 0$
n - 10 = $0$, n - 6 = $0 \Rightarrow n = 10, 6$
We get two values of 'n' as sum of $7^{th}$ term to $10^{th}$ term is zero as some terms are negative and some are positive.

$S_6 = 36 \Rightarrow \frac{6}{2}[2a + 5d] = 36$
$2a + 5d = 12$ ----- (1)
$S_{16} = 256 \Rightarrow \frac{16}{2}[2a + 15d] = 256$
$2a +15d = 32$ ----- (2)
Solving (1) and (2)
$d = 2$
$a = 1$
$S_{10} = \frac{10}{2}[2(1) +9(2)]$
$= 100$

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

(D) 4

(B) $12$

(c) $23$

(c) $-250$

(C) $45$

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)

$S_n = 4n^2 - n$
$a_{25}= S_{25} - S_{24}$ (1/2 Mark)
$= [4(25)^2-25] - [4(24)^2 – 24]$ (1 Mark)
$= 195$ (1/2 Mark)

$a = 15, S_{15} = 750$
$\frac{15}{2}[2a + 14d] = 750$
$2(15) + 14d = 100$
$d = 5$
$a_{20} = a + 19d = 15 + 19(5) = 110$

$a_p = a + (p - 1)d = q$
quad (i)
$a_q = a + (q - 1)d = p$
quad (ii)
Solving (i) and (ii)
$d = -1, a = q + p - 1$
$a_n = (q + p - 1) + (n - 1)(-1) = q + p - n$

Here $S_n = 6n - n^2$
For $n = 1$, $S_1 = a_1 = 6(1) - (1)^2 = 6 - 1 = 5$
So, the first term $a = 5$.
For $n = 2$, $S_2 = a_1 + a_2 = 6(2) - (2)^2 = 12 - 4 = 8$
We know $S_2 = a_1 + a_2 = a + (a+d) = 2a+d$.
So, $2a+d = 8$.
Substitute $a=5$: $2(5) + d = 8$
$10 + d = 8$
$d = 8 - 10 = -2$
Now, find the $30^{th}$ term $a_{30}$:
$a_{30} = a + (30-1)d$
$a_{30} = 5 + 29(-2)$
$a_{30} = 5 - 58$
$a_{30} = -53$

$a = 8, l = 65$
$$\begin{aligned}& 730 = \frac{n}{2}[8+65] \\ & n = \frac{730 \times 2}{73} = 20\end{aligned}$$
$$\begin{aligned}& \therefore l = a + 19d \implies 65 = 8 + 19d \\ & \implies d = 3\end{aligned}$$

$S_m = S_n$
$\Rightarrow \frac{m}{2}[2a + (m - 1)d] = \frac{n}{2}[2a + (n - 1)d]$
$\Rightarrow 2a(m - n) = d(n^2 - m^2) - d(n - m)$
$\Rightarrow 2a = -d(m + n - 1)$
or $2a + (m + n - 1)d = 0$
i.e., $S_{m+n} = \frac{m+n}{2}[2a + (m + n - 1)d] = 0$

Let $a$ be the first term and $d$ be the common difference.
$\frac{a + 9d}{a + 29d} = \frac{1}{3}$
$\Rightarrow a=d \dots (i)$
$\frac{6}{2}(2a + 5d) = 42$
$\Rightarrow 2a + 5d = 14 \dots (ii)$
Solving (i) and (ii)
$a=2$ and $d=2$

$\frac{14}{2}(20 + 13d) = 1050$
$\Rightarrow d = 10$
$\therefore a_{20} = 10 + 19 \times 10 = 200$
$a_n = 10 + (n - 1) 10 = 10n$

$a = 5, a_n = 45, S_n = 400$
$\frac{n}{2}(5 + 45) = 400$
$\Rightarrow n = 16$
$5+15d=45$
$\Rightarrow d = \frac{40}{15} \text{ or } \frac{8}{3}$

Let first term = $a$ and common difference = $d$
$(a + 14d) = (a + 7d) + 21$ (1 Mark)
$\Rightarrow d = 3$ (1/2 Mark)
Also, $S_{10} = 55 = \frac{10}{2}[2a + 9 \times 3]$ (1/2 Mark)
$\Rightarrow a = -8$ (1/2 Mark)
$\therefore$ A. P. is $-8, -5, -2, ...$ (1/2 Mark)

$S_n = 2n^2 + 13n$
$S_1 = a_1 = 15$ (I Mark)
$S_2 = a_1 + a_2 = 34 \Rightarrow a_2 = 19$ (II Mark)
$\Rightarrow d = 19 - 15 = 4$ (III Mark)
$\therefore a_n = 15 + (n - 1) \times 4 = 4n + 11$ (IV Mark)
Hence $a_{10} = 4 \times 10 + 11 = 51$ (V Mark)

$a = 7, d = \frac{21}{2} - 7 = \frac{7}{2}$ [$1$ mark]
$84 = 7 + (n - 1) \times \frac{7}{2} \Rightarrow n = 23$ [$1$ mark]
$S_{23} = \frac{23}{2}(7 + 84) = \frac{2093}{2}$ [$1$ mark]

$S_1 = a = 5$ [$\frac{1}{2}$ mark]
$S_2 = 12$ [$\frac{1}{2}$ mark]
Therefore $d = 2$ [$1$ mark]
Hence $a_{15} = 33$ [$1$ 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)

$a_n = 5 + 2n \Rightarrow a = 7, d = 2$ (1 mark)
$S_{20} = \frac{20}{2} [14 + 19 \times 2] = 520$ (1 mark)
$52 = 7 + (n - 1) \times 2 \Rightarrow n = \frac{47}{2}$, which is not a natural number. ($\frac{1}{2}$ mark)
Therefore, 52 cannot be a term of this A.P. ($\frac{1}{2}$ mark)

This is an Arithmetic Progression (AP).
$a = -4$, $d = -1 - (-4) = 3$.
Let $x$ be the $n^{th}$ term, so $a_n = x$.
Sum of $n$ terms $S_n = 437$.
$S_n = \frac{n}{2}[2a + (n-1)d]$
$437 = \frac{n}{2}[2(-4) + (n-1)3]$
$437 = \frac{n}{2}[-8 + 3n - 3]$
$437 = \frac{n}{2}[3n - 11]$
$874 = 3n^2 - 11n$
$3n^2 - 11n - 874 = 0$
Using quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $n = \frac{11 \pm \sqrt{(-11)^2 - 4(3)(-874)}}{2(3)}$
$n = \frac{11 \pm \sqrt{121 + 10488}}{6} = \frac{11 \pm \sqrt{10609}}{6} = \frac{11 \pm 103}{6}$
$n = \frac{11+103}{6} = \frac{114}{6} = 19$ or $n = \frac{11-103}{6} = \frac{-92}{6}$ (not possible as $n$ must be positive integer).
So, $n = 19$.
Now find $x = a_n = a_{19}$.
$a_n = a + (n-1)d$
$x = -4 + (19-1)3 = -4 + 18(3) = -4 + 54 = 50$.
So, $x = 50$.

Given $S_n = 5n^2 + 3n$.
For $n=1$, $S_1 = 5(1)^2 + 3(1) = 5+3 = 8$. So, $a_1 = 8$.
For $n=2$, $S_2 = 5(2)^2 + 3(2) = 5(4) + 6 = 20+6 = 26$.
$S_2 = a_1 + a_2 \Rightarrow 26 = 8 + a_2 \Rightarrow a_2 = 18$.
Common difference $d = a_2 - a_1 = 18 - 8 = 10$.
Now, $a_n = 168$.
$a_n = a_1 + (n-1)d$
$168 = 8 + (n-1)10$
$160 = (n-1)10$
$16 = n-1 \Rightarrow n = 17$.
To find $20^{th}$ term, $a_{20} = a_1 + (20-1)d = 8 + 19(10) = 8 + 190 = 198$.

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, $S_5:S_{21}=25:189$

$1+4+7+10 + \dots + x = 287$
$a = 1, d = 3$ Last term $= x$
$\Rightarrow 1 + (n - 1) 3 = x$
$3n - 2 = x \Rightarrow n = \frac{x + 2}{3}$
dots (i)
$S_n = 287$
$\frac{n}{2} [1 + x] = 287 \Rightarrow \frac{(\frac{x + 2}{3})}{2} [1 + x] = 287$
quad using (i)
$\frac{(x + 2)(x + 1)}{6} = 287$
$(x + 2)(x + 1) = 1722$
$x^2 + 3x + 2 = 1722$
$x^2 + 3x - 1720 = 0$
$(x + 43)(x - 40) = 0$
$\Rightarrow x = -43, 40$
$x \ne -43$ (as terms are increasing positive numbers)
$\therefore x = 40$

$\frac{a + 10d}{a + 17d} = \frac{2}{3}$
$3a +30d = 2a + 34d \Rightarrow a = 4d$
Therefore, $\frac{a + 4d}{a + 20d} = \frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \frac{1}{3}$
$\frac{S_5}{S_{21}} = \frac{\frac{5}{2}[2a + 4d]}{\frac{21}{2}[2a +20d]} = \frac{5[8d + 4d]}{21[8d + 20d]}$
$= \frac{5 \times 12d}{21 \times 28d} = \frac{5}{49}$ or $S_5: S_{21} = 5:49$

$-4+(-1)+2 + 5 + ..... + x = 437$
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 $a$ be the first term and $d$ be the common difference.
$S_7 = 182 \Rightarrow \frac{7}{2}(2a + 6d) = 182$
$\Rightarrow 2a + 6d = \frac{182 \times 2}{7} = 52$
$a + 3d = 26$
quad ----- (i)
$\frac{a_4}{a_{17}} = \frac{1}{5} \Rightarrow \frac{a + 3d}{a + 16d} = \frac{1}{5}$
$\Rightarrow 5a + 15d = a+16d$
$4a = d$
quad --------(ii)
Solving (i) and (ii)
$a = 2$ and $d=8$
$\therefore \text{AP is } 2, 10, 18, 26, ....$

$S_q = 63q - 3q^2$
$\therefore S_1 = 60 \Rightarrow a_1 = 60$ ($1^{\text{st}}$ term)
$S_2 = 63(2) - 3(2)^2 = 126 - 12 = 114$
$a_1 + a_2 = 114 \Rightarrow a_2 = 114 - 60 = 54$
$d = a_2 - a_1 = 54 - 60 = -6$
$a_p = -60$
$60 + (p-1)d = -60$
$(p-1)(-6) = -120 \Rightarrow p=21$
$a_{11} = a + 10d = 60 + 10(-6) = 0$

$a + a_8 = 32 \Rightarrow 2a + 7d = 32$ ----- (i)
$a \times a_8 = 60 \Rightarrow a(a + 7d) = 60$ ----- (ii)
Solving (i) & (ii), we get
$a=2$ or $a = 30$
and $d = 4$ or $d = -4$
First term and common difference of A.P. are $2$ and $4$ or $30$ and $-4$ respectively.

Here $n = 40$,
$S_9 = \frac{9}{2} [2a +8d] = 153 \Rightarrow a + 4d = 17$ ----- (i)
and $S_{40} - S_{34} = 687$ or $a_{35} + a_{36} + a_{37} + a_{38} + a_{39} + a_{40} = 687$
$\Rightarrow 6a+219d = 687$ or $2a + 73d = 229$ ----- (ii)
solving (i) and (ii) to get $a = 5, d = 3$
Also, $S_{40} = \frac{40}{2} (10 + 39 \times 3) = 2540$

Let first term $= a$ and common difference $= d$
ATQ, $(a + 2d) + (a + 6d) = 6$
$a + 4d = 3$
$a = 3-4d$
Also, $(a + 2d)(a + 6d) = 8$
$(3 - 4d + 2d)(3 - 4d + 6d) = 8$
$9 - 4d^2 = 8$
$d = \pm \frac{1}{2}$
When $d = \frac{1}{2} \Rightarrow a = 1$
$S_{16} = \frac{16}{2}[2 \times 1 + 15 \times \frac{1}{2}]$
$= 76$
When $d = -\frac{1}{2} \Rightarrow a = 5$
$S_{16} = \frac{16}{2}[2 \times 5 + 15 \times (-\frac{1}{2})]$
$= 20$

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

Let first term = $a$ and common difference = $$\begin{aligned}& d \\ & text{ATQ, } \frac{a_{10}}{a_{30}} = \frac{a + 9d}{a + 29d} = \frac{1}{3} \\ & 3a + 27d = a + 29d \\ & 2a = 2d \Rightarrow a = d \\ & S_6 = \frac{6}{2}[2a + (6-1)d] = 42 \\ & 3[2a + 5a] = 42 \quad (\text{since } a=d) \\ & 3[7a] = 42 \\ & 21a = 42 \\ & a = 2 \\ & d = 2 \\ & a_{13} = a + 12d = 2 + 12 \times 2 = 2 + 24 = 26\end{aligned}$$

(i)
$a + 2d = 4$ ..... (i) (1/2 Mark)
$a + 8d = -8$ ..... (ii) (1/2 Mark)
solving (i) and (ii) to get $d = -2$, $a = 8$ (1/2 + 1/2 Mark)
$a_n = 0 \Rightarrow 8 + (n - 1) (-2) = 0$ (1 Mark)
$\Rightarrow n = 5$ i.e. $5^{th}$ term is zero.
(ii) $S_n = -36 \Rightarrow \frac{n}{2}[16 + (n - 1) (-2)] = -36$ (1 Mark)
$\Rightarrow n^2 - 9n - 36 = 0$
$\Rightarrow (n - 12) (n + 3) = 0 \Rightarrow n = 12$ (rejecting $n = -3$) (1/2 + 1/2 Mark)

(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)

$56, 63, ..., 497$
Here $a = 56$ and $d=7$
Let $a_n = 497$
$\Rightarrow 56 + (n-1) \times 7 = 497$
$\Rightarrow n = 64$
$S_{64} = \frac{64}{2} \times (56 + 497) = 17696$

$11, 15, ..., 299$
Here $a = 11$ and $d=4$
Let $a_n = 299$
$\Rightarrow 11 + (n-1) \times 4 = 299$
$\Rightarrow n = 73$
$S_{73} = \frac{73}{2} \times (11 +299) = 11315$

A.P formed is $1000, 1100, 1200, ...$
$a = 1000, d = 100$
$a_{30} = a + 29d = 3900$
Amount paid in $30^{th}$ instalment $= Rs \quad 3,900$
$S_{30} = \frac{30}{2}[2 \times 1000 + 29 \times 100] = 15 \times 4900 = 73,500$
$\therefore$ Total amount paid after $30^{th}$ instalment $= Rs\quad 73,500$

Let the numbers be $a - d, a, a + d$
$\therefore a - d + a + a + d = 24$
$\Rightarrow a = 8$
Also, $(a - d)^2 + a^2 + (a + d)^2 = 194$
$\Rightarrow (8 - d)^2 + 8^2 + (8 + d)^2 = 194$
$\Rightarrow d^2 = 1 \Rightarrow d = \pm 1$
$\therefore$ Numbers are $7, 8, 9$ or $9,8,7$

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 $= 140 + 26 \times 140$
$= \text{Rs}3780$

$3$ - digit natural numbers divisible by $11$ are
$110, 121, ..., 990$ ($1/2$)
Here first term $= 110$ and common difference $= 11$
$a_n = 990$
$\Rightarrow 110 + (n - 1) \times 11 = 990$ ($1$)
$\Rightarrow n = 81$ ($1/2$)
$S_{81} = \frac{81}{2} \times [110+990]$
$= 44550$ ($1$)

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) $51, 49, 47, 45, \dots, 31$
(ii) Here $a = 51 \& d = -2$
$31 = 51 + (n - 1) (-2)$
$\Rightarrow n = 11$
So, minimum $11$ days he need to practice to achieve the goal.
(iii) (a) $a_n = 51 + (n - 1) (-2)$
$a_n = 53 - 2n$
OR
(b) $21 = 51 + (n - 1) (-2)$
$n = 16$
So, minimum $16$ days he need to practice to achieve the goal.

(i) Number on the first spot $$\begin{aligned}& = 20 + 4 \times 1 = 24 \\ & \text{(ii) (a) } 20 + 4n = 112 \\ & \Rightarrow n = 23\ \text{OR} \\ & \text{(ii) (b) } d = 4 \\ & S_{10} = \frac{10}{2} [2 \times 24 + 9 \times 4] \\ & = 420 \\ & \text{(iii) Number on the } (n - 2)^{th} \text{ spot } = 20 + 4(n - 2) \\ & = 12 + 4n\end{aligned}$$

(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) $$\begin{aligned}& S_6 = \frac{6}{2} [2 \times 250 + 5 \times 110] \\ & = 3150\end{aligned}$$

(i) A.P. formed is
$104, 117, 130, \dots$ with $a = 104$ and $d = 13$
$a_7 = 104 + 6 \times 13 = 182$
(ii) $a_{14} - a_9 = 5 \times 13 = 65$
(iii) (a) Last term of A.P. is $988$ and $d = -13$
$a_9 = 988 + 8 \times (-13) = 884$
OR
(b) Last term of A.P. is $988$ and $d = -13$
$988 = 104 + (n - 1) \times 13$
$\Rightarrow n = 69$
$S_{69} = \frac{69}{2} \times (104+988)$
$= 37674$

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

AP formed is $200, 350, 500, ...$
(i) Distance of $10^{th}$ pole from base = $A_{10}$
$= 200+9\times 150$
$= 1550$ m
(ii) Distance between $15^{th}$ pole and $25^{th}$ pole = $a_{25} - A_{15}$
$= 10 \times 150 = 1500$ m
(iii) (a) Distance of $15^{th}$ pole from the top = $300 + 14 \times 150$
$= 2400$ m
Time taken by cable car = $\frac{2400}{5} = 480$ seconds or $8$ minutes
OR
(iii) (b) Distance of last pole from the base = $(5000 – 300)$ m = $4700$ m
$\therefore a_n = 4700$
$\Rightarrow 200+ (n – 1)150 = 4700$
Solving, we get $n = 31$

Given A.P. is 1, 3, 5, ...
(i) $a_{10} = 1 + 9 \times 2 = 19$
(ii) $a_4$ (from bottom) $= 19 + 3 \times (-2) = 13$
(iii) (a) $S_8 = \frac{8}{2} \times [2 \times 1 + 7 \times 2] = 64$
(iii) (b) Number of triangles from $5^{th}$ row to $10^{th}$ row $= S_{10} - S_4 = \frac{10}{2} \times [2 \times 1 + 9 \times 2] - \frac{4}{2} \times [2 \times 1 + 3 \times 2] = 84$. Number of triangles in first 4 rows, $S_4 = \frac{4}{2} \times [2 \times 1 + 3 \times 2] = 16$. Required number of triangles $= 84 - 16 = 68$

Here AP is 400, 407.6, 415.2, ... (i) $a_6 = 400 + 5(7.6) = 438$ m (1 mark). (ii) $a_8 - a_4 = 30.4$ m (1 mark). (iii) $S_6 = \frac{6}{2}(2 \times 400 + 5 \times 7.6) = 2514$ m (1 + 1 marks). OR (iii) Total distance covered = $S_8 - S_3 = \frac{8}{2}(2 \times 400 + 7 \times 7.6) - \frac{3}{2}(2 \times 400 + 2 \times 7.6) = 2190$ m (1 + 1 marks).

(a) $a_5 = 1 + 4(1) = 5$
(b) $S_7 = \frac{7}{2} [2 + 6 \times 1] = 28$ coins
(c) (i) $S_n = \frac{n}{2} [2 + (n - 1) \times 1] = 300 \implies n = 24$ weeks
OR
(c) (ii) $S_{20} = 10[20 + 19 \times 10] = \text{\text{Rs} } 2100$

(i) Required distance for the first potato $= 5 + 5 = 10$ m (I) (1 Mark)
(ii) Required distance for the second potato $= 8 + 8 = 16$ m (I) (1 Mark)
(iii) (a) Distances covered form an A.P. with $a = 10$ and $d = 6$ (I) (1 Mark)
$\therefore S_{10} = \frac{10}{2} [2 \times 10 + 9 \times 6]$
$= 5 \times 74 = 370$ m (II) (1 Mark)
OR
(iii) (b) Distances covered form an A.P. with $a = 10$ and $d = 6$ (I) (1/2 Mark)
$\therefore S_{10} = \frac{10}{2} [2 \times 10 + 9 \times 6]$
$= 5 \times 74 = 370$ m (II) (1 Mark)
Time $= \frac{370}{5} = 74$ seconds (III) (1/2 Mark)

$a = 1000, d = 100$
(i) $a_{30} = 1000 + 29(100)$ (1 Mark)
$= 3900$
Amount paid in $30^{th}$ instalment = ₹ $3900$
(ii) $a_{40} = 1000 + 39(100)$ (1 Mark)
$= 4900$
Amount paid in last instalment = ₹ $4900$
(iii) (a) $S_{30} = \frac{30}{2} \times [2(1000) + 29 \times 100]$ (1 Mark)
$= 73500$
Amount still he has to pay = $118000 – 73500 = \text{Rs}44500$ (1 Mark)
OR
(iii) (b) $\frac{a_{10}}{a_{40}} = \frac{1000+900}{1000+3900}$ (1 Mark)
$= \frac{1900}{4900}$
$= \frac{19}{49}$ (1 Mark)
The ratio is $19:49$

(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

(i) $60, 57, 54, 51, 48$ (0.5 Mark)
This is an A.P. as common difference is $(-3)$ (0.5 Mark)
(ii) Required time = $45$ seconds (1 Mark)
(iii) (a) $a_n = 60 + (n-1) (-3) = 30$ (1 Mark)
$\Rightarrow n = 11$ (1 Mark)
OR
(b) $a_n = 60 + (n-1) (-3.2) = 28$ (1 Mark)
$\Rightarrow n = 11$ (1 Mark)

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 = $a_{20} = 22 + 19 (- 1) = 3$

ATQ,
$32+36 +40 + ..... = 2,000$
$a = 32, d =4$ ($\frac{1}{2} + \frac{1}{2}$)
Let 'n' be the number of months.
$S_n = 2000$
$\frac{n}{2}[2(32) + (n - 1) 4] = 2000$ (1)
$n^2 + 15n - 1000 = 0$ (1)
$n^2 + 40n - 25n - 1000 = 0$
$(n + 40) (n - 25) = 0$ (1)
$\Rightarrow n = -40, n = 25$
(rejecting)
$\therefore n = 25$ (1)
So, Prerna will save ₹2,000, in $25$ months.

(i) Integers divisible by $9$ are $108, 117, 126, ....., 198$
a = $108$, d = $9$
a + $(n-1)d = 198$
$\Rightarrow 108 + (n - 1)9 = 198 \Rightarrow n = 11$
$S_{11} = \frac{n}{2}(a+l) = \frac{11}{2}(108 + 198)$
$= 1683$
(ii) Sum of all integers = $\frac{99}{2}(101 + 199) = \frac{99}{2} \times 300 = 14850$
Sum of integers not divisible by $9 = 14850 - 1683 = 13167$

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) $\frac{19}{12}$