NCERT Solutions: Class 10 Maths Chapter 5 - Arithmetic Progressions

A list is an AP only when consecutive differences are equal. In (i), each new fare adds ` 8. In (iii), each new metre adds ` 50. In (ii) and (iv), the next amount is obtained by multiplying by a fixed factor, not by adding a fixed number.

The first four terms are $a, a+d, a+2d, a+3d$. Substituting the given $a$ and $d$ gives the listed terms.

The first term is the first listed number. The common difference is found from second term minus first term.

For each list, compare consecutive differences. Equal differences give an AP, and the next three terms are found by adding that common difference repeatedly.

Use $a_n = a + (n - 1)d$. For example, in (ii), $0 = -18 + 9d$, so $d = 2$. In (iv), $3.6 = -18.9 + (n - 1)2.5$, so $n = 10$.

1. A. As listed in the question.
2. B. As listed in the question.
3. C. As listed in the question.
4. D. As listed in the question.

(i) Here $a = 10$, $d = -3$. So $a_{30} = 10 + 29(-3) = -77$.
(ii) Here $a = -3$, $d = \dfrac{5}{2}$. So $a_{11} = -3 + 10 \times \dfrac{5}{2} = 22$.

Use equal spacing between the given terms. For (iii), $d = \dfrac{9\dfrac{1}{2} - 5}{3} = \dfrac{3}{2}$, giving $5, \dfrac{13}{2}, 8, \dfrac{19}{2}$. For (v), $a_2 = 38$ and $a_6 = -22$, so $4d = -60$ and $d = -15$.

Here $a = 3$, $d = 5$. Let $a_n = 78$. Then $78 = 3 + (n - 1)5$, so $75 = 5(n - 1)$ and $n = 16$.

(i) $205 = 7 + (n - 1)6$, so $n = 34$.
(ii) $-47 = 18 + (n - 1)(-\dfrac{5}{2})$, so $n - 1 = 26$ and $n = 27$.

Here $a = 11$, $d = -3$. If $-150$ were a term, $-150 = 11 + (n - 1)(-3)$, giving $n - 1 = \dfrac{161}{3}$, not an integer. Hence it is not a term.

$a_{16} - a_{11} = 5d = 73 - 38 = 35$, so $d = 7$. Then $a_{31} = a_{11} + 20d = 38 + 140 = 178$.

$a_3 = a + 2d = 12$ and $a_{50} = a + 49d = 106$. Subtracting gives $47d = 94$, so $d = 2$ and $a = 8$. Hence $a_{29} = 8 + 28 \times 2 = 64$.

$a_3 = a + 2d = 4$ and $a_9 = a + 8d = -8$. Subtracting gives $6d = -12$, so $d = -2$ and $a = 8$. Now $0 = 8 + (n - 1)(-2)$, so $n = 5$.

$a_{17} - a_{10} = (a + 16d) - (a + 9d) = 7d$. Given $7d = 7$, so $d = 1$.

The common difference is $12$. Since $132 = 11 \times 12$, the required term is 11 places after the 54th term: $54 + 11 = 65$.

Let the first terms be $a$ and $b$, with the same common difference $d$. The difference between nth terms is $[a + (n - 1)d] - [b + (n - 1)d] = a - b$, independent of $n$. Therefore it remains 100.

The first three-digit multiple of 7 is 105 and the last is 994. These form an AP with $d = 7$. So $994 = 105 + (n - 1)7$, giving $n = 128$.

The multiples are $12, 16, 20, \ldots, 248$. Thus $248 = 12 + (n - 1)4$, so $n = 60$.

For the first AP, $a_n = 63 + (n - 1)2 = 2n + 61$. For the second AP, $a_n = 3 + (n - 1)7 = 7n - 4$. Equating, $2n + 61 = 7n - 4$, so $n = 13$.

$a_3 = a + 2d = 16$. Also, $a_7 - a_5 = (a + 6d) - (a + 4d) = 2d = 12$, so $d = 6$. Then $a + 12 = 16$, giving $a = 4$.

Read the AP backwards from 253 with common difference $-5$. The 20th term from the last is $253 + 19(-5) = 158$.

$(a + 3d) + (a + 7d) = 24$, so $a + 5d = 12$. Also, $(a + 5d) + (a + 9d) = 44$, so $a + 7d = 22$. Subtracting gives $2d = 10$, hence $d = 5$ and $a = -13$.

The salaries form an AP with $a = 5000$ and $d = 200$. Let ` 7000 be the nth year salary: $7000 = 5000 + (n - 1)200$, so $n = 11$. Counting 1995 as the first year, the 11th year is 2005.

Weekly savings form an AP with $a = 5$ and $d = 1.75$. Thus $20.75 = 5 + (n - 1)1.75$, so $n - 1 = 9$ and $n = 10$.

Use $S_n = \dfrac{n}{2}[2a + (n - 1)d]$. Substituting the respective values of $a$, $d$ and $n$ gives the listed sums.

Use $S_n = \dfrac{n}{2}(a+l)$ after finding $n$. For (i), $a=7$, $d=\dfrac{7}{2}$, $l=84$, so $n=23$ and $S=\dfrac{23}{2}(91)=\dfrac{2093}{2}$. The same formula gives (ii) $286$ and (iii) $-8930$.

Apply $a_n = a + (n - 1)d$ and $S_n = \dfrac{n}{2}[2a + (n - 1)d]$ or $S_n = \dfrac{n}{2}(a+l)$ as appropriate. For instance, in (iv), $a+2d=15$ and $5(2a+9d)=125$, giving $d=-1$ and then $a_{10}=8$.

Here $a=9$, $d=8$. Then $636 = \dfrac{n}{2}[18 + (n-1)8] = n(4n+5)$. So $4n^2 + 5n - 636 = 0$, giving $n = 12$ as the positive solution.

$400 = \dfrac{n}{2}(5+45) = 25n$, so $n = 16$. Then $45 = 5 + 15d$, so $d = \dfrac{40}{15} = \dfrac{8}{3}$.

$350 = 17 + (n - 1)9$, so $n - 1 = 37$ and $n = 38$. Then $S = \dfrac{38}{2}(17+350)=19\times367=6973$.

$a_{22}=a+21d=149$, so $a+147=149$ and $a=2$. Therefore $S_{22}=\dfrac{22}{2}(2+149)=1661$.

$d = 18 - 14 = 4$ and $a = 14 - 4 = 10$. Hence $S_{51}=\dfrac{51}{2}[2(10)+50(4)]=\dfrac{51}{2}\times220=5610$.

$S_7=49$ gives $\dfrac{7}{2}(2a+6d)=49$, so $a+3d=7$. $S_{17}=289$ gives $\dfrac{17}{2}(2a+16d)=289$, so $a+8d=17$. Therefore $d=2$ and $a=1$. Thus $S_n=\dfrac{n}{2}[2+2(n-1)]=n^2$.

(i) $a_1=7$, $a_2=11$, so the common difference is $4$; $a_{15}=63$ and $S_{15}=\dfrac{15}{2}(7+63)=525$.
(ii) $a_1=4$, $a_2=-1$, so the common difference is $-5$; $a_{15}=-66$ and $S_{15}=\dfrac{15}{2}(4-66)=-465$.

$S_n = 4n - n^2$. So $S_1=3$ and $S_2=8-4=4$. The second term is $S_2-S_1=1$. Also, $a_3=S_3-S_2=(12-9)-4=-1$ and $a_{10}=S_{10}-S_9=(40-100)-(36-81)=-15$. In general, $a_n=S_n-S_{n-1}=5-2n$.

The first 40 positive integers divisible by 6 are $6, 12, 18, \ldots, 240$. Their sum is $S=\dfrac{40}{2}(6+240)=4920$.

The multiples are $8, 16, 24, \ldots, 120$. Hence $S=\dfrac{15}{2}(8+120)=960$.

The odd numbers are $1, 3, 5, \ldots, 49$, with 25 terms. Therefore $S=\dfrac{25}{2}(1+49)=625$.

The penalties form an AP with $a=200$, $d=50$, $n=30$. Thus $S_{30}=\dfrac{30}{2}[2(200)+29(50)]=15(1850)=27750$.

Let the largest prize be $a$. Then $d=-20$, $n=7$, and $700=\dfrac{7}{2}[2a+6(-20)]$. So $2a-120=200$, giving $a=160$. The seven prizes are then found by subtracting 20 each time.

For one section in Classes I to XII, the number of trees is $1+2+\cdots+12 = \dfrac{12}{2}(1+12)=78$. There are 3 sections of each class, so total trees $=3\times78=234$.

Length of a semicircle is $\pi r$. The radii are $0.5, 1.0, 1.5, \ldots, 6.5$ for 13 semicircles. Their sum is $\dfrac{13}{2}(0.5+6.5)=45.5$. Total length $=\pi \times 45.5 = \dfrac{22}{7}\times\dfrac{91}{2}=143$ cm.

Rows form an AP with $a=20$, $d=-1$, and sum 200. Thus $200=\dfrac{n}{2}[40+(n-1)(-1)]=\dfrac{n}{2}(41-n)$. This gives $n^2-41n+400=0$, so $n=16$ or $25$. The value $25$ is not possible because it would make the top row negative, so $n=16$. Top row $=20-15=5$.

One-way distances to the potatoes are $5, 8, 11, \ldots, 32$ for 10 potatoes. Each potato requires a round trip, so total distance is $2\times\dfrac{10}{2}(5+32)=370$ m.

Here $a=121$, $d=-4$. For the first negative term, $121+(n-1)(-4) \lt 0$, i.e. $125-4n \lt 0$. Thus $n \gt 31.25$, so the least natural number is $32$.

$a_3+a_7=6$ gives $(a+2d)+(a+6d)=6$, so $a+4d=3$. Also, $(a+2d)(a+6d)=8$. These two terms are $3-2d$ and $3+2d$, so $9-4d^2=8$, giving $d=\pm\dfrac{1}{2}$. If $d=\dfrac{1}{2}$, then $a=1$ and $S_{16}=76$. If $d=-\dfrac{1}{2}$, then $a=5$ and $S_{16}=20$.

The distance between top and bottom rungs is $2\dfrac{1}{2}$ m = 250 cm. With rungs 25 cm apart, there are $\dfrac{250}{25}+1=11$ rungs. Their lengths form an AP from 45 cm to 25 cm, so total length $=\dfrac{11}{2}(45+25)=385$ cm.

Sum before house $x$ is $1+2+\cdots+(x-1)=\dfrac{x(x-1)}{2}$. Sum after it is $(x+1)+\cdots+49=\dfrac{49\times50}{2}-\dfrac{x(x+1)}{2}$. Equating, $x(x-1)=2450-x(x+1)$, so $2x^2=2450$, $x^2=1225$, and $x=35$.

The volume corresponding to the first step is $\dfrac{1}{4}\times\dfrac{1}{2}\times50=\dfrac{25}{4}\text{ m}^3$. The terrace uses volumes $\dfrac{25}{4}, 2\times\dfrac{25}{4}, \ldots, 15\times\dfrac{25}{4}$. Thus total volume $=\dfrac{25}{4}(1+2+\cdots+15)=\dfrac{25}{4}\times120=750\text{ m}^3$.