Class 10 Maths: Arithmetic Progressions – Extra Questions with Answers

Ans: C

Solution:
The sequence is an AP.
First term (a) = 2
Common difference (d) = $\frac{5}{2} – 2 = \frac{1}{2}$
Next term = $\frac{7}{2} + \frac{1}{2} = 4$

Answer: a = 2, d = 1/2, next term = 4

Ans: A

Solution:
Since $x-2$, $6$, $x+2$ are in AP, the difference between consecutive terms is constant.
Therefore, $6 – (x-2) = (x+2) – 6$.
Simplifying, $6-x+2 = x+2-6$.
$8-x = x-4$.
$2x = 12$.
$x=6$.
Answer: 6

Ans: A

Solution:
$\sum_{r=1}^{5} (4 + 3^r) = \sum_{r=1}^{5} 4 + \sum_{r=1}^{5} 3^r = 4 \cdot 5 + (3^1 + 3^2 + 3^3 + 3^4 + 3^5) = 20 + (3 + 9 + 27 + 81 + 243) = 20 + 363 = 383$
Answer: 383

Ans: B

Solution:
$S_1 = 3(1)^2 + 2(1) = 5 = a_1$
$S_2 = 3(2)^2 + 2(2) = 16$
$a_2 = S_2 – S_1 = 16 – 5 = 11$
$d = a_2 – a_1 = 11 – 5 = 6$
$a_3 = a_2 + d = 11 + 6 = 17$

Answer: 5, 11, 17

Ans: A

Solution:
2 hours and 30 minutes is equal to 5 increments of 30 minutes. The penalty amounts form an arithmetic progression: 1000, 1100, 1200, 1300, 1400. The sum of an arithmetic series is given by: S = (n/2) * (2a + (n-1)d) where n is the number of terms, a is the first term, and d is the common difference. In this case, n=5, a=1000, and d=100. S = (5/2) * (2*1000 + (5-1)*100) = (5/2) * (2000 + 400) = (5/2) * 2400 = 6000.

Answer: 6000

Ans: A

Solution:
Let $a_n$ be the $n$th term of the first A.P. and $b_n$ be the $n$th term of the second A.P.
For the first A.P., the first term is $a_1 = 5$ and the common difference is $d_1 = 13 – 5 = 8$. So, $a_n = a_1 + (n-1)d_1 = 5 + (n-1)8 = 5 + 8n – 8 = 8n – 3$.
For the second A.P., the first term is $b_1 = 40$ and the common difference is $d_2 = 43 – 40 = 3$. So, $b_n = b_1 + (n-1)d_2 = 40 + (n-1)3 = 40 + 3n – 3 = 3n + 37$.
We want to find $n$ such that $a_n = b_n$.
Therefore, $8n – 3 = 3n + 37$.
$8n – 3n = 37 + 3$
$5n = 40$
$n = 8$

Answer: 8

Ans: A

Solution:
Since $a, b, c$ are in AP, $2b = a+c$.
Since $b, c, d$ are in GP, $c^2 = bd$.
Since $c, d, e$ are in HP, $\frac{2}{d} = \frac{1}{c} + \frac{1}{e}$.

Given $a=3$ and $e=12$, we have $2b = 3+c$, $c^2 = bd$, and $\frac{2}{d} = \frac{1}{c} + \frac{1}{12}$.
From $2b = 3+c$, we get $b = \frac{3+c}{2}$.
From $\frac{2}{d} = \frac{1}{c} + \frac{1}{12} = \frac{12+c}{12c}$, we have $d = \frac{24c}{12+c}$.
Substituting $b$ and $d$ into $c^2 = bd$, we have $c^2 = \frac{3+c}{2} \cdot \frac{24c}{12+c}$.
So $c^2 = \frac{12c(3+c)}{12+c}$.
Since $c \neq 0$, we can divide by $c$: $c = \frac{12(3+c)}{12+c}$.
$c(12+c) = 36 + 12c$
$12c + c^2 = 36 + 12c$
$c^2 = 36$
$c = \pm 6$.

If $c=6$, then $b=\frac{3+6}{2} = \frac{9}{2}$ and $d = \frac{24 \cdot 6}{12+6} = \frac{144}{18} = 8$.
If $c=-6$, then $b=\frac{3-6}{2} = -\frac{3}{2}$ and $d = \frac{24 \cdot (-6)}{12-6} = \frac{-144}{6} = -24$.

When $c=6$, $b, c, d$ are in GP: $\frac{9}{2}, 6, 8$.
$\frac{6}{9/2} = \frac{12}{9} = \frac{4}{3}$ and $\frac{8}{6} = \frac{4}{3}$.
$a, b, c$ are in AP: $3, \frac{9}{2}, 6$. $\frac{9}{2}-3 = \frac{3}{2}$ and $6-\frac{9}{2} = \frac{3}{2}$.
When $c=-6$, $b, c, d$ are in GP: $-\frac{3}{2}, -6, -24$.
$\frac{-6}{-3/2} = 4$ and $\frac{-24}{-6} = 4$.
$a, b, c$ are in AP: $3, -\frac{3}{2}, -6$.
$-\frac{3}{2} – 3 = -\frac{9}{2}$ and $-6-(-\frac{3}{2}) = -6+\frac{3}{2} = -\frac{9}{2}$.

Since $a, b, c$ are in AP and $b, c, d$ are in GP and $c, d, e$ are in HP, and $e=12$, we can conclude that $c>0$, so we choose $c=6$.

Answer: $b=\frac{9}{2}, c=6, d=8$

Ans: C

Solution:
Let $n$ be the number of terms.
The first term $a = 5$, and the common difference $d = 11 – 5 = 6$.
The sum of an A.P. is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
We are given $S_n = 455$.
So, $455 = \frac{n}{2}[2(5) + (n-1)6]$
$910 = n[10 + 6n – 6]$
$910 = n(4 + 6n)$
$910 = 4n + 6n^2$
$6n^2 + 4n – 910 = 0$
$3n^2 + 2n – 455 = 0$
Using the quadratic formula,
$n = \frac{-2 \pm \sqrt{2^2 – 4(3)(-455)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 5460}}{6} = \frac{-2 \pm \sqrt{5464}}{6} = \frac{-2 \pm 74}{6}$
Since $n$ must be positive, $n = \frac{-2 + 74}{6} = \frac{72}{6} = 12$.

Answer: 12

Ans: B

Solution:
Since $x_1, x_2, x_3, \dots$ are in A.P., let the common difference be $d$.
We have $x_1 + x_7 + x_{10} = x_1 + (x_1 + 6d) + (x_1 + 9d) = 3x_1 + 15d = -9$.
Dividing by 3, we get $x_1 + 5d = -3$, which means $x_6 = -3$.
We also have $x_3 + x_8 + x_{12} = (x_1 + 2d) + (x_1 + 7d) + (x_1 + 11d) = 3x_1 + 20d = -16$.
Now, we want to find $x_3 + x_8 + x_{22} = (x_1 + 2d) + (x_1 + 7d) + (x_1 + 21d) = 3x_1 + 30d = 3(x_1 + 10d)$.
We have $3x_1 + 15d = -9$, and $3x_1 + 20d = -16$. Subtracting these, we get $5d = -7$, so $d = -\frac{7}{5}$.
Then $x_1 + 5d = x_1 + 5(-\frac{7}{5}) = x_1 – 7 = -3$, so $x_1 = 4$.
Now $x_3 + x_8 + x_{22} = 3x_1 + 30d = 3(4) + 30(-\frac{7}{5}) = 12 – 42 = -30$.

Alternatively, we know that $3x_1 + 15d = -9$ and $3x_1 + 20d = -16$.
Subtracting, we get $-5d = 7$, so $d = -7/5$.
Now we calculate $3x_1 + 30d = 3x_1 + 15d + 15d = -9 + 15d = -9 + 15(-7/5) = -9 – 21 = -30$.
Thus $x_3 + x_8 + x_{22} = 3x_1 + 30d = -30$.

Answer: -30

Ans: B

Solution:
Let the first term be ‘a’ and the common difference be ‘d’.
7th term: a + 6d
3rd term: a + 2d
Given: a + 6d = 2(a + 2d) => a + 6d = 2a + 4d => a – 2d = 0 … (1)
10th term: a + 9d
5th term: a + 4d
Given: a + 9d = (1/4)(a + 4d) + 2 => 4a + 36d = a + 4d + 8 => 3a + 32d = 8 … (2)
From (1), a = 2d. Substituting in (2): 3(2d) + 32d = 8 => 6d + 32d = 8 => 38d = 8 => d = 4/19
a = 2d = 8/19
13th term: a + 12d = 8/19 + 12(4/19) = 8/19 + 48/19 = 56/19

Answer: 56/19