# Grade 10 – Math – Arithmetic Progression- Important Questions and Answers

Question 1: The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

Solution:

We can write the given condition as:

Cost of digging a well for first metre = Rs.150
Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200
Cost of digging a well for first 3 metres = Rs.200+50 = Rs.250
Cost of digging a well for first 4 metres =Rs.250+50 = Rs.300
And so on..

So, 150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each term.

Question 2: If $3 x + k + 1,2 x + 10$ and $x + 14$ a c. K is three consecutive terms, find the value of $k$.

Solution:
\begin {aligned} & (2 x + 10) – (3 x + k + 1) = (x + 14) – (2 x + 10) \\ & \Rightarrow -x-k + 9 = -x + 4 \\ & \Rightarrow -k + 9 = 4 \\ & \Rightarrow k = 5 \end {aligned}

Question 3: Which of the following are APs?
(i) 3, 3 + √2, 3 + 2√2, 3 + 3√2….
(ii) aa2a3a4 …
(iii) √3, √6, √9, √12 …
(iv) 12, 52, 72, 73 …

Solution:

(i) Given, 3, 3+√2, 3+2√2, 3+3√2

Here,
a2 – a1 = 3+√2-3 = √2
a3 – a2 = (3+2√2)-(3+√2) = √2
a4 – a3 = (3+3√2) – (3+2√2) = √2

Since, an+1 – an or the common difference is same every time , it’s an AP.

(ii) aa2a3a4 …

Here,
a2 – a1 = a2a = a(a-1)
a3 – a2 = aaa2(a-1)
a4 – a3 = a4 – aa3(a-1)

Since, an+1 – an or the common difference is not same every time, this is not an AP.

(iii) √3, √6, √9, √12 …

Here,
a2 – a1 = √6-√3 = √3×√2-√3 = √3(√2-1)
a3 – a2 = √9-√6 = 3-√6 = √3(√3-√2)
a4 – a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)

Since, an+1 – an or the common difference is not same every time.

Therefore, the given series doesn’t form a A.P.

(iv) 12, 52, 72, 73 …

Or 1, 25, 49, 73 …

Here,
a2 − a1 = 25−1 = 24
a3 − a= 49−25 = 24
a4 − a3 = 73−49 = 24

Since, an+1 – an or the common difference is same every time.

Question 4: $a_{n}=n^{2}+1, x \in N$ is a series. Prove that this is not an AP.

Solution :
\begin{aligned} & a_{n}=n^{2}+1 \\ & \text { Providing, } n=1,2,3,4, \ldots \\ &a_{1}=1^{2}+1=1+1=2 \\ &a_{2}=2^{2}+1=4+1=5 \\ &a_{3}=3^{2}+1=9+1=10\\ &a_{4}=4^{2}+1=16+1=17\\ \\ & \text{ So,the series is,} 2,5,10,17, \ldots \ldots \ldots \\ \\ & a_{2}-a_{1}=5-2=3 \\ & a_{3}-a_{2}=10-5=5 \\ & \because a_{2}-a_{1} \neq a_{3}-a_{2} \\ &\therefore \text{ The series is not an AP.} \end{aligned}

Question 5: Taxi fare after each kilometer is like, while the fare for the first kilometer is $15$ and the fare for each additional kilometer is Rs. 8. Is this an AP series?

Solution: First kilometer fare = 15 rupees.
Additional kilometer fare = 8 rupees
Series is $: 15,23,31,39 \ldots \ldots$
Check:
$a = 15$
${d} _ {1} ={a} _ {2} – {a} _ {1} = 23-15 = 8$
${d} _ {2} = {a} _ {3} – {a} _ {2} = 31-23 = 8$
${d} _ {3} = {a} _ {4} – {a} _ {3} = 39-31 = 8$
Since the difference of all differences is the same, that is, the difference = 8. Hence the given list is A. P.

Questions 1. A person started working in a firm in 1995 at a salary of ₹ 10000 per annum. He received ₹ 500 as salary increment per year. In which year his salary reached ₹ 15000.

Solution: $\begin {array} {l} a = 10,000 \\ d = 500 \\ l = 15000 \end {array}$ Suppose his salary reached ₹ 15000 in $n$ years. Then
$\begin {array} {lr} \ l = a + (n-1) d\\ \Rightarrow 15000 = 10000 + (n + 1) 500 \\ \Rightarrow (n-1) 500 = 15000-10000\\ \Rightarrow (n-1) 500 = 5000 \Rightarrow n-1 = \frac {5000} {50} \Rightarrow n = 10 \\ \Rightarrow n = 11 \\ 1995 + 11 = 2006 \end{array}$
Hence the salary of that person reached ₹ 15000 in the year 2006.

Question 2. The 4th term of an A.P. is zero. Prove that the 25 th term of the A.P. is three times its 11 th term.
Solution:
Let $a$ be first term and $d$ be the common difference of the A.P. Then,
\begin{aligned} & a_{n} =a+(n-1) d \\ & a_{4} =a+(4-1) d \\ & 0 =a+3 d \\ & \Rightarrow a=-3d \quad \because \text{ Given } a_4=0 \end{aligned}
Now,
\begin{aligned} &a_{25} =a+(25-1) d \\ &=a+24 d=-3 d+24 d=21 d=3 \times 7 d \end{aligned}
Hence,
\begin{aligned} &a_{25}=3 \times a_{11} & \\ &\left[\because \text { since } a_{11}=a+(11-1) d=-3 d+10 d=7 d\right] \end{aligned}

Question 3 : The digits of a positive number of three digits are in A.P. and their sum is $15 .$ The number obtained by
reversing the digits is 594 less than the original number. Find the number.
Solution:
Let the required numbers in A.P. are $a-d, a, a+d$ respectively.
$\begin{array}{l} \text { Now, } \\ a-d+a+a+d=15 {[\because \text { Sum of digits }=15]} \\ \Rightarrow 3 a=15 \Rightarrow a=5 \end{array}$
According to question, number is $100(a-d)+10 a+a+d,$ i.e. $111 a-99 d$
Number on reversing the digits is $100(a+d)+10 a+a-d,$ i.e. $111 a+99 d$
Now, as per given condition in question,
\begin{aligned} (111 a-99 d)-(111 a+99 d) &=594 \\ -198 d &=594 \\ d &=-3 \end{aligned}
$\therefore \quad$ Digits of number are $[5-(-3), 5,(5+(-3)]=8,5,2$
$\therefore \quad$ Required number is $111 \times(5)-99(-3)=555+297=852$

Question 4: Balls are arranged in rows to form an equilateral triangle .The first row consists of one ball, the second two balls and so on. If 669 more balls are added, then all the balls can be arranged in the shape of a square and each of its sides then contains 8 balls less than each side of the triangle. find the initial number of balls.

Answer: Let their be n balls in each side of the triangle $\therefore$ No. of ball (in $\Delta)=1+2+3 \ldots \ldots \ldots .=\frac{n(n+1)}{2}$
No. of balls in each side square $=n-8$
No. of balls in square $=(n-8)^{2}$ $\mathrm{APQ} \frac{n(n+1)}{2}+660=(\mathrm{n}-8)^{2}$
On solving
$n^{2}+n+1320=2\left(n^{2}-16 n+64\right)$
$n^{2}-33 n-1210=0$
$\Rightarrow(n-55)(n+22)=0$
$n=-22(N \cdot P)$
$n=55$
$\therefore$ No. of balls $=\frac{n(n+1)}{2}=\frac{55 \times 56}{2}$

Question 5: How many numbers of two digit are divisible by 3?

Solution:The first 2 digit number divisible by 3 is 12. And, the last 2 digit number divisible by 3 is 99.

So, this forms an A.P.

12, 15, 18, 21, …. , 99

Where, a = 12 and d = 3

Finding the number of terms in this A.P

⇒ 99 = 12 + (n-1)3
99 = 12 + 3n – 3
90 = 3n
n = 90/3 = 30

Therefore, there are 30 two digit numbers divisible by 3.

Question 1: If the term $n$ is $(2 n-1)$ of an AP, find the sum of its first n terms.

Solution:
\begin {aligned} &a_{n} = 2 n-1 \\ &n = 1,2, \text {if placed} \\ &a_ {1} = 2 (1) -1 = 2-1 = 1 \\ &a_ {2} = 2 (2) -1 = 4-1 = 3 \\ &a = a_ {1} = 1 \\ &d = a_ {2} -a_ {1} = 3-1 = 2\\ &\therefore \text{ Sum of first n terms }\\ &= \frac {n} {2} [2 a + (n-1) d]\\ &= \frac {n} {2} [2 (1) + (n-1) 2] \\ &= n ^ {2} \end {aligned}

Question 2: Find the sum of first 25 terms of an A.P whose nth term is given by an = 7 – 3n.

Solution:

Given an A.P. whose nth term is given by an = 7 – 3n

To find the sum of the n terms of the given A.P., we use the formula,

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

Where, a = the first term , l = the last term.

Putting n = 1 in the given an, we get

a = 7 – 3(1) = 7 – 3 = 4

For the last term (l), here n = 25

a15 = 7 – 3(25) = -68

So, Sn = 25(4 – 68)/2

= 25 x (-32)

= -800

Therefore, the sum of the 15 terms of the given A.P. is S25 = -800

Question 3:
Find the middle term of the sequence formed by all three-digit numbers which leave a remainder 5 when divided by 7. Also find the sum of all numbers on both sides of the middle term separately.

Solution:
List of three digit number that leaves a remainder of $5,$ when divided by 7 are $103,110,117,$ $\ldots 999$
These numbers are in AP with $a=103, d=7, a_{n}=999,$ where $n=$ number of terms
\begin{aligned} & \Rightarrow a+(n-1) d=999 \\ &\Rightarrow 103+(n-1) 7=999 \Rightarrow(n-1) 7=896\\ &\Rightarrow n-1=128 \\ &\Rightarrow n=129\\ & \text{since number of terms is odd, so only one middle term}\\ & \therefore \text { Middle term }=\frac{129+1}{2}=65 \mathrm{th}=\left(\frac{n+1}{2}\right)^{\text {th }}\\ &a_{65}=a+64 d=103+64 \times 7=103+448=551 ] \end{aligned}

Number of terms before 65 th term $=64$
\begin{aligned} &\therefore S_{64} =\frac{64}{2}(2 \times 103+63 \times 7) & \{\because {S}_{n}=\frac{n}{2}(2 a+(n-1) d)\}\\ &=32(206+441)=20704 \\ &a_{129} =999=a+128 d\\ &\therefore \quad S_{129}=\frac{129}{2}\left[a+a_{129}\right]=71079 \end{aligned}
Now, sum of terms after middle term $={S}_{129}-\left({S}_{64}+551\right)=49824$

Question 4:
In a school, students decided to plant 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 double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students. Which value is shown in this question?

Solution:
According to question, each section of :
Class I will plant 2 trees, class II will plant 4 trees, class III will plant 6 trees and so on.. class 12 will plant 24 trees and each class has 2 sections.
$\therefore$ Number of trees planted $=4+8+12+\cdots+48$
This forms an A.P. with $a=4, d=4$ and $n=12$
$\therefore$ Number of trees planted,
$S_{12}=\frac{12}{2}(4+48)=6 \times 52=312 \quad \quad \{\because S_{n}=\frac{n}{2}[2 a+(n-1) d]\}$
Students are concerned about safety and pollution free environment.

Question 5: Find the sum of all multiples of 9 lying between 400 and 800 .

Solution:
Multiples of 9 between 400 and 800 are: $405,414,423, \ldots, 792$ Here,
$a=405 ; d=9 ; \text { last term } a_{n}=792$
$\begin{array}{lrl} \text { Now, } a_{n}=792 & \\ \Rightarrow a+(n-1) d=792 \Rightarrow 405-9(n+1)=792 \\ \Rightarrow 9(n-1)=387 \Rightarrow n-1=43 \\ \Rightarrow n=44 \end{array}$
Now, sum of these multiples,
${S}{44}=\frac{44}{2}(405+792) \quad\left[\because {S}{n}=\frac{n}{2}\left(a_{1}+a_{n}\right)\right]\\ =22 \times 1197=26334$