$\implies n = \frac{-(5) \pm \sqrt{(5)^2 – 4(4)(-636)}}{2(4)}$

$\implies n = \frac{-5 \pm \sqrt{25 + 10176}}{8}$

$\implies n = \frac{-5 \pm \sqrt{10201}}{8}$

$\implies n = \frac{-5 \pm 101}{8}$

Either,

$n = \frac{-5 + 101}{8} = \frac{96}{8} = 12$

or,

$n = \frac{-5 – 101}{8} = \frac{-106}{8} = -\frac{53}{4}$ (Not admissible as the number of terms cannot be negative or a fraction)

$\therefore n = 12$

Therefore, 12 terms of the AP must be taken to give a sum of 636.

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

$\mathbf{Sol^n.}$

Given:

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

We know that:

$S_n = \frac{n}{2}(a + l)$

$\implies 400 = \frac{n}{2}(5 + 45)$

$\implies 400 = \frac{n}{2}(50)$

$\implies 400 = 25n$

$\implies n = \frac{400}{25}$

$\implies n = 16$

Again, we know,

$l = a_n = a + (n – 1)d$:

$45 = 5 + (16 – 1)d$

$\implies 45 – 5 = 15d$

$\implies 40 = 15d$

$\implies d = \frac{40}{15}$

$\implies d = \frac{8}{3}$

$\therefore$ The number of terms $n = \mathbf{16}$ and the common difference $d = \mathbf{\frac{8}{3}}$.

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

$\mathbf{Sol^n.}$

Given:

$a = 17$, $l = a_n = 350$, and $d = 9$

We know that:

$a_n = a + (n – 1)d$

$\implies 350 = 17 + (n – 1)9$

$\implies 350 – 17 = 9(n – 1)$

$\implies 333 = 9(n – 1)$

$\implies n – 1 = \frac{333}{9}$

$\implies n – 1 = 37$

$\implies n = 37 + 1$

$\implies n = 38$

Again, we know,

$S_n = \frac{n}{2}(a + l)$

$\implies S_{38} = \frac{38}{2}(17 + 350)$

$= 19 \times 367$

$= 6973$

$\therefore$ There are 38 terms and their sum is 6973.

7. Find the sum of first 22 terms of an AP in which $d = 7$ and $22^{\text{nd}}$ term is 149.

$\mathbf{Sol^n.}$

Given:

$d = 7$, $n = 22$, and $a_{22} = 149$

We know that:

$a_n = a + (n – 1)d$

$\implies a_{22} = a + 21d$

$\implies 149 = a + 21(7)$

$\implies 149 = a + 147$

$\implies a = 149 – 147$

$\implies a = 2$

Again, we know,

$S_n = \frac{n}{2}(a + l)$

$S_{22} = \frac{n}{2}(a + l)$

$= \frac{22}{2}(2 + 149)$

$= 11 \times 151$

$= 1661$

$\therefore$ The sum of the first 22 terms of the AP is 1661.

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

$\mathbf{Sol^n.}$

Given:

Second term, $a_2 = 14$

Third term, $a_3 = 18$

∴ Common difference,

$d = a_3 – a_2 = 18 – 14 = 4$

Now, the first term:

$a = a_2 – d = 14 – 4 = 10$

We need to find $S_{51}$ where $n = 51$;

We know,

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$\implies S_{51} = \frac{51}{2}[2(10) + (51 – 1)4]$

$= \frac{51}{2}[20 + 50(4)]$

$= \frac{51}{2}[20 + 200]$

$= \frac{51}{2} \times 220$

$= 51 \times 110$

$= 5610$

$\therefore$ The sum of the first 51 terms is 5610.

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first $n$ terms.

$\mathbf{Sol^n.}$

Given:

$S_7 = 49$

$\implies \frac{7}{2}[2a + (7 – 1)d] = 49$

$\implies \frac{7}{2}[2a + 6d] = 49$

$\implies 7(a + 3d) = 49$

$\implies a + 3d = \frac{49}{7}$

$\implies a + 3d = 7 \quad \text{— (1)}$

Also given:

$S_{17} = 289$

$\implies \frac{17}{2}[2a + (17 – 1)d] = 289$

$\implies \frac{17}{2}[2a + 16d] = 289$

$\implies 17(a + 8d) = 289$

$\implies a + 8d = \frac{289}{17}$

$\implies a + 8d = 17 \quad \text{— (2)}$

Subtracting (1) from (2), we have

$(a + 8d) – (a + 3d) = 17 – 7$

$\implies 5d = 10$

$\implies d = \frac{10}{5}$

$\implies d = 2$

Substituting $d = 2$ in equation (1), we ahve

$a + 3(2) = 7$

$\implies a + 6 = 7$

$\implies a = 7 – 6$

$\implies a = 1$

Now, finding the sum of the first $n$ terms:

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$= \frac{n}{2}[2(1) + (n – 1)2]$

$= \frac{n}{2}[2 + 2n – 2]$

$= \frac{n}{2}(2n)$

$= n^2$

$\therefore$ The sum of the first $n$ terms is $n^2$.

10. Show that $a_1, a_2, \dots, a_n, \dots$ form an AP where $a_n$ is defined as below:

(i) $a_n = 3 + 4n$

(ii) $a_n = 9 – 5n$

Also find the sum of the first 15 terms in each case.

(i) $a_n = 3 + 4n$

$\mathbf{Sol^n.}$

Given:

$a_n = 3 + 4n$

Putting $n = 1, 2, 3, 4, \dots$:

$a_1 = 3 + 4(1) = 3 + 4 = 7$

$a_2 = 3 + 4(2) = 3 + 8 = 11$

$a_3 = 3 + 4(3) = 3 + 12 = 15$

$a_4 = 3 + 4(4) = 3 + 16 = 19$

Thus, $a_1, a_2, \dots, a_n, \dots$ is:

$7, 11, 15, 19, \dots$

Now, we have

$a_2 – a_1 = 11 – 7 = 4$

$a_3 – a_2 = 15 – 11 = 4$

$a_4 – a_3 = 19 – 15 = 4$

i.e.,$a_{k+1} – a_k$ is the same every time.

Therefore, $7, 11, 15, 19, \dots$ forms an AP with first term $a = 7$ and common difference $d = 4$.

Now, the sum of the first 15 terms ($n = 15$):

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$\implies S_{15} = \frac{15}{2}[2(7) + (15 – 1)4]$

$= \frac{15}{2}[14 + 14 \times 4]$

$= \frac{15}{2}[14 + 56]$

$= \frac{15}{2} \times 70$

$= 15 \times 35$

$= 525$

$\therefore$ The sum of the first 15 terms is 525.

(ii) $a_n = 9 – 5n$

$\mathbf{Sol^n.}$

Given:

$a_n = 9 – 5n$

Putting $n = 1, 2, 3, 4, \dots$:

$a_1 = 9 – 5(1) = 9 – 5 = 4$

$a_2 = 9 – 5(2) = 9 – 10 = -1$

$a_3 = 9 – 5(3) = 9 – 15 = -6$

$a_4 = 9 – 5(4) = 9 – 20 = -11$

The list of numbers is:

$4, -1, -6, -11, \dots$

Now, we have,

$a_2 – a_1 = -1 – 4 = -5$

$a_3 – a_2 = -6 – (-1) = -6 + 1 = -5$

$a_4 – a_3 = -11 – (-6) = -11 + 6 = -5$

Since $a_{k+1} – a_k$ is constant, $4, -1, -6, -11, \dots$ forms an AP with first term $a = 4$ and common difference $d = -5$.

Now, the sum of the first 15 terms ($n = 15$):

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$\implies S_{15} = \frac{15}{2}[2(4) + (15 – 1)(-5)]$

$= \frac{15}{2}[8 + 14(-5)]$

$= \frac{15}{2}[8 – 70]$

$= \frac{15}{2} \times (-62)$

$= 15 \times (-31)$

$= -465$

$\therefore$ The sum of the first 15 terms is -465.

11. If the sum of the first $n$ terms of an AP is $4n – n^2$, what is the first term (that is $S_1$)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the $n$th terms.

$\mathbf{Sol^n.}$

Given: $S_n = 4n – n^2$

For the first term ($S_1$):

Putting $n = 1$:

$S_1 = 4(1) – (1)^2$

$= 4 – 1$

$= 3$

$\therefore$ First term, $a = a_1 = S_1 = 3$

For the sum of first two terms ($S_2$):

Putting $n = 2$:

$S_2 = 4(2) – (2)^2$

$= 8 – 4$

$= 4$

$\therefore$ The sum of the first two terms is 4.

For the second term ($a_2$):

We know that, $a_2 = S_2 – S_1$

$\implies a_2 = 4 – 3 = 1$

$\therefore$ The second term is 1.

Now, common difference:

$d = a_2 – a_1 = 1 – 3 = -2$

For the 3rd term ($a_3$):

$a_3 = a + 2d$

$= 3 + 2(-2)$

$= 3 – 4$

$= -1$

For the 10th term ($a_{10}$):

$a_{10} = a + 9d$

$= 3 + 9(-2)$

$= 3 – 18$

$= -15$

For the $n$th term ($a_n$):

$a_n = a + (n – 1)d$

$= 3 + (n – 1)(-2)$

$= 3 – 2n + 2$

$= 5 – 2n$

$\therefore$ First term is 3, sum of first two terms is 4, second term is 1, 3rd term is -1, 10th term is -15, and the $n$th term is $5 – 2n$.

12. Find the sum of the first 40 positive integers divisible by 6.

$\mathbf{Sol^n.}$

The first 40 positive integers divisible by 6 are:

$6, 12, 18, 24, \dots$

Clearly the given numbers are in AP with, $a = 6$, $d = 12 – 6 = 6$, and $n = 40$

We know that:

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$\implies S_{40} = \frac{40}{2}[2(6) + (40 – 1)6]$

$= 20[12 + 39 \times 6]$

$= 20[12 + 234]$

$= 20 \times 246$

$= 4920$

$\therefore$ The sum of the first 40 positive integers divisible by 6 is 4920.

13. Find the sum of the first 15 multiples of 8.

$\mathbf{Sol^n.}$

The first 15 multiples of 8 are:

$8, 16, 24, 32, \dots$

Clearly the given numbers are in AP with, $a = 8$, $d = 16 – 8 = 8$, and $n = 15$

We know that:

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$\implies S_{15} = \frac{15}{2}[2(8) + (15 – 1)8]$

$= \frac{15}{2}[16 + 14 \times 8]$

$= \frac{15}{2}[16 + 112]$

$= \frac{15}{2} \times 128$

$= 15 \times 64$

$= 960$

$\therefore$ The sum of the first 15 multiples of 8 is 960.

14. Find the sum of the odd numbers between 0 and 50.

$\mathbf{Sol^n.}$

The odd numbers between 0 and 50 are:

$1, 3, 5, 7, \dots, 49$

Clearly the given numbers are in AP with, $a = 1$, $d = 3 – 1 = 2$, and $l = a_n = 49$

We know that:

$a_n = a + (n – 1)d$

$\implies 49 = 1 + (n – 1)2$

$\implies 49 – 1 = 2(n – 1)$

$\implies 48 = 2(n – 1)$

$\implies n – 1 = \frac{48}{2}$

$\implies n – 1 = 24$

$\implies n = 24 + 1$

$\implies n = 25$

Again, we know,

$S_n = \frac{n}{2}(a + l)$

$\implies S_{25} = \frac{25}{2}(1 + 49)$

$= \frac{25}{2} \times 50$

$= 25 \times 25$

$= 625$

$\therefore$ The sum of the odd numbers between 0 and 50 is 625.

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

$\mathbf{Sol^n.}$

The penalties for successive days form an AP:

$200, 250, 300, \dots$

Where, $a = 200$, $d = 250 – 200 = 50$, and $n = 30$

We know that:

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$\implies S_{30} = \frac{30}{2}[2(200) + (30 – 1)50]$

$= 15[400 + 29 \times 50]$

$= 15[400 + 1450]$

$= 15 \times 1850$

$= 27750$

$\therefore$ The contractor has to pay Rs 27,750 as penalty.

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

$\mathbf{Sol^n.}$

Let the value of the first prize be Rs $a$.

Since each prize is Rs 20 less than its preceding prize, $d = -20$.

Total number of prizes, $n = 7$

Total sum of prizes, $S_n = 700$

We know that:

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$\implies 700 = \frac{7}{2}[2a + (7 – 1)(-20)]$

$\implies \frac{700 \times 2}{7} = 2a + 6(-20)$

$\implies 100 \times 2 = 2a – 120$

$\implies 200 = 2a – 120$

$\implies 2a = 200 + 120$

$\implies 2a = 320$

$\implies a = \frac{320}{2}$

$\implies a = 160$

Now, the values of each of the 7 prizes are:

• 1st prize $= a = \text{Rs } 160$

• 2nd prize $= 160 – 20 = \text{Rs } 140$

• 3rd prize $= 140 – 20 = \text{Rs } 120$

• 4th prize $= 120 – 20 = \text{Rs } 100$

• 5th prize $= 100 – 20 = \text{Rs } 80$

• 6th prize $= 80 – 20 = \text{Rs } 60$

• 7th prize $= 60 – 20 = \text{Rs } 40$

$\therefore$ The value of each of the prizes are Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

$\mathbf{Sol^n.}$

Number of classes $= 12$

Each class has 3 sections.

Number of trees planted by:

Class I $= 3 \times 1 = 3$

Class II $= 3 \times 2 = 6$

Class III $= 3 \times 3 = 9$

$\dots$

Class XII $= 3 \times 12 = 36$

The series forms an AP:

$3, 6, 9, \dots, 36$

Here, $a = 3$, $d = 6 – 3 = 3$, $n = 12$, and $l = 36$

We know that:

$S_n = \frac{n}{2}(a + l)$

$\implies S_{12} = \frac{12}{2}(3 + 36)$

$= 6 \times 39$

$= 234$

$\therefore$ A total of 234 trees will be planted by the students.

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … as shown in Fig. 5.4. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi = \frac{22}{7}$)

$\mathbf{Sol^n.}$

We know that the perimeter of a semicircle of radius $r$ is given by $l = \pi r$.

Radii of successive semicircles are:

$r_1 = 0.5\text{ cm}, \; r_2 = 1.0\text{ cm}, \; r_3 = 1.5\text{ cm}, \; r_4 = 2.0\text{ cm}, \dots$

Length of successive semicircles:

$l_1 = \pi(0.5)$

$l_2 = \pi(1.0)$

$l_3 = \pi(1.5)$

$\dots$$\dots$
$\dots$$\dots$

These lengths form an AP where:

$a = 0.5\pi$

$d = 1.0\pi – 0.5\pi = 0.5\pi$

Number of semicircles, $n = 13$

Total length of the spiral, $S_{13}$:

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$\implies S_{13} = \frac{13}{2}[2(0.5\pi) + (13 – 1)(0.5\pi)]$

$= \frac{13}{2}[\pi + 12(0.5\pi)]$

$= \frac{13}{2}[\pi + 6\pi]$

$= \frac{13}{2} \times 7\pi$

$= \frac{13}{2} \times 7 \times \frac{22}{7}$[Taking $\pi = \frac{22}{7}$]

$= 13 \times 11$

$= 143\text{ cm}$

$\therefore$ The total length of the spiral made up of 13 consecutive semicircles is 143 cm.

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row?

$\mathbf{Sol^n.}$

The number of logs in successive rows forms an AP:

$20, 19, 18, \dots$

Here, $a = 20$, $d = 19 – 20 = -1$, and $S_n = 200$

We know that:

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$\implies 200 = \frac{n}{2}[2(20) + (n – 1)(-1)]$

$\implies 400 = n[40 – n + 1]$

$\implies 400 = n[41 – n]$

$\implies 400 = 41n – n^2$

$\implies n^2 – 41n + 400 = 0$

$\implies n^2 – 25n – 16n + 400 = 0$

$\implies n(n – 25) – 16(n – 25) = 0$

$\implies (n – 25)(n – 16) = 0$

Either $n – 25 = 0 $
$\implies n = 25$

or $n – 16 = 0 $
$\implies n = 16$

For $n = 25$ rows,

$a_{25} = a + 24d $
$= 20 + 24(-1) $
$= 20 – 24 $
$= -4$ [ Not admissible as the number of logs cannot be negative]

For$n = 16$ rows,

$a_{16} = a + 15d $
$= 20 + 15(-1) $
$= 20 – 15 $
$= 5$

$\therefore$ The 200 logs are placed in 16 rows, and there are 5 logs in the top row.

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6).

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

$\mathbf{Sol^n.}$

Distance run by the competitor to pick up and drop:

• 1st potato $= 2 \times 5 = 10\text{ m}$

• 2nd potato $= 2 \times (5 + 3) = 2 \times 8 = 16\text{ m}$

• 3rd potato $= 2 \times (5 + 3 + 3) = 2 \times 11 = 22\text{ m}$

Clearly, these distances form an AP:

$10, 16, 22, \dots$

Here, $a = 10$, $d = 16 – 10 = 6$, and total potatoes $n = 10$

We know that:

$S_n = \frac{n}{2}[2a + (n – 1)d]$

∴ The total distance the competitor has to run is:

$\implies S_{10} = \frac{10}{2}[2(10) + (10 – 1)6]$

$= 5[20 + 9 \times 6]$

$= 5[20 + 54]$

$= 5 \times 74$

$= 370\text{ m}$

$\therefore$ The total distance the competitor has to run is 370 m.

21. The first and the last terms of an AP are 1 and 11 respectively. If the sum of its terms is 36, then the number of terms will be

(A) 5

(B) 6

(C) 7

(D) 8

$\mathbf{Sol^n.}$

Given:

$a = 1$, $l = 11$, and $S_n = 36$

We know that:

$S_n = \frac{n}{2}(a + l)$

$\implies 36 = \frac{n}{2}(1 + 11)$

$\implies 36 = \frac{n}{2}(12)$

$\implies 36 = 6n$

$\implies n = \frac{36}{6}$

$\implies n = 6$

Correct option is (B) 6

22. Match the column

Choose the correct option:

Correct Answer: C) P-3, Q-4, R-1, S-2

$\mathbf{Sol^n.}$

• P) The sum of first $n$ natural numbers is given by $\frac{n(n+1)}{2} \rightarrow \mathbf{3}$

• Q) The sum upto $n$ terms of an AP is $\frac{n}{2}\{2a+(n-1)d\} \rightarrow \mathbf{4}$

• R) The sum of first $n$ odd natural numbers is $n^2 \rightarrow \mathbf{1}$

• S) The sum of first $n$ even natural numbers is $n(n+1) \rightarrow \mathbf{2}$

Therefore, the matching order is P-3, Q-4, R-1, S-2.

Correct option is C)

23. Which of the following is correct for the AP 5, 9, 13, …, 185?

(i) $a_n = 1 + 4n$

(ii) $a_3 + a_4 = 30$

(iii) $S_n = \frac{n}{2} \times 190$

(iv) $n = 43$

A) (i), (iii)

B) (ii), (iii)

C) (i), (ii), (iii)

D) (i), (ii), (iii), (iv)

Correct Answer: C) (i), (ii), (iii)

$\mathbf{Sol^n.}$

Given AP: $5, 9, 13, \dots, 185$

Here, $a = 5$, $d = 9 – 5 = 4$, and $l = a_n = 185$

Checking statement (i):

$a_n = a + (n – 1)d$

$= 5 + (n – 1)4$

$= 5 + 4n – 4$

$= 1 + 4n$ (True)

Checking statement (ii):

$a_3 = 13$

$a_4 = 13 + 4 = 17$

$a_3 + a_4 = 13 + 17 = 30$ (True)

Checking statement (iii):

$S_n = \frac{n}{2}(a + l)$

$= \frac{n}{2}(5 + 185)$

$= \frac{n}{2} \times 190$ (True)

Checking statement (iv):

$a_n = 185$

$\implies 1 + 4n = 185$

$\implies 4n = 184$

$\implies n = \frac{184}{4} = 46 \ne 43$ (False)

Therefore, statements (i), (ii), and (iii) are correct.

Correct option is C) (i), (ii), (iii)

24. Assertion (A): The sum of first 10 terms of the sequence 11, 13, 15, 17, …….. is 200.

Reason (R): The sum of first $n$ odd natural number is $n^2$

(A) (A) and (R) both are true and (R) is the correct explanation for (A)

(B) (A) and (R) both are true but (R) is not the correct explanation for (A)

(C) (A) is true but (R) is false

(D) (A) is false but (R) is true.

Correct option is: (B) 

$\mathbf{Sol^n.}$

For Assertion (A):

Sequence: $11, 13, 15, 17, \dots$

Here, $a = 11$, $d = 13 – 11 = 2$, and $n = 10$

$S_{10} = \frac{10}{2}[2(11) + (10 – 1)2]$

$= 5[22 + 9(2)]$

$= 5[22 + 18]$

$= 5 \times 40$

$= 200$

Hence, Assertion (A) is true.

For Reason (R):

Sum of the first $n$ odd natural numbers is indeed $n^2$ (True).

However, the sequence in Assertion (A) starts from 11, not 1, so the formula in Reason (R) is not the direct method used to calculate (A).

Correct option is (B) (A) and (R) both are true but (R) is not the correct explanation for (A)

25. If the sum of the first $n$ odd number is $S_1$ and the sum of the first $n$ natural number is $S_2$, then the value of $\frac{S_1}{S_2}$ is

(A) $\frac{2n}{n+1}$

(B) $\frac{n+1}{2n}$

(C) $\frac{n}{n+1}$

(D) $\frac{n+1}{n}$

$\mathbf{Sol^n.}$

We know that:

Sum of first $n$ odd numbers, $S_1 = n^2$

Sum of first $n$ natural numbers, $S_2 = \frac{n(n + 1)}{2}$

Now,

$\frac{S_1}{S_2} = \frac{n^2}{\frac{n(n + 1)}{2}}$

$= \frac{2n^2}{n(n + 1)}$

$= \frac{2n}{n + 1}$

Correct option is (A) $\frac{2n}{n+1}$

26. A fruit seller brings a cartoon of apples which contains 240 apples. He arranges the apples such that in the first row 6 apples, in the second row there are 10 apples, in the third row there are 14 apples and so on.

(i) Find the total number of rows of apples

(ii) Find the difference in number of apples placed in 7th and 3rd row.

(iii) Find the number of apples in last 3rd row

(iv) Check if there is any row containing 32 apples.

$\mathbf{Sol^n.}$

The arrangement of apples forms an AP:

$6, 10, 14, \dots$

Here, $a = 6$, $d = 10 – 6 = 4$, and total apples $S_n = 240$

(i) Total number of rows of apples:

We know that:

$S_n = \frac{n}{2}[2a + (n – 1)d]$

$\implies 240 = \frac{n}{2}[2(6) + (n – 1)4]$

$\implies 240 = \frac{n}{2}[12 + 4n – 4]$

$\implies 240 = \frac{n}{2}[4n + 8]$

$\implies 240 = n(2n + 4)$

$\implies 240 = 2n^2 + 4n$

$\implies 2n^2 + 4n – 240 = 0$

$\implies n^2 + 2n – 120 = 0$

$\implies n^2 + 12n – 10n – 120 = 0$

$\implies n(n + 12) – 10(n + 12) = 0$

$\implies (n + 12)(n – 10) = 0$

Either $n – 10 = 0 $
$\implies n = 10$

or $n + 12 = 0 $
$\implies n = -12$ (Not admissible, as the number of rows cannot be negative)

$\therefore$ Total number of rows of apples is 10.

(ii) Find the difference in number of apples placed in 7th and 3rd row.

Number of apples placed in 7th row: $a_7 = (a + 6d)$

Number of apples placed in 3rd row: $a_3 =  (a + 2d)$

Now the difference in number of apples placed in 7th and 3rd row:

$a_7 – a_3 = (a + 6d) – (a + 2d)$

$= 4d$

$= 4 \times 4$

$= 16$

$\therefore$ The difference is 16.

(iii) Number of apples in last 3rd row:

Since total rows $n = 10$, the last 3rd row is the 8th row.

∴ $a_8 = a + 7d$

$= 6 + 7(4)$

$= 6 + 28$

$= 34$

$\therefore$ Number of apples in the last 3rd row is 34.

(iv) Check if there is any row containing 32 apples:

Let the $n$th row contain 32 apples:

$a_n = 32$

$\implies a + (n – 1)d = 32$

$\implies 6 + (n – 1)4 = 32$

$\implies 4(n – 1) = 32 – 6$

$\implies 4(n – 1) = 26$

$\implies n – 1 = \frac{26}{4}$

$\implies n – 1 = \frac{13}{2} = 6.5$

$\implies n = 7.5$

Since $n$ must be a positive integer, there is no row containing 32 apples.