Apply to Aakash iACST Scholarship Test 2024
NCERT Solutions for Exercise 5.3 Class 10 Maths Chapter 5 Arithmetic Progressions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. The sum of n terms formula of the Arithmetic Progression is used to solve 20 problems in class 10 ex 5.3. There are also some word problems to help with comprehension. In exercise 5.3 Class 10 Maths, a number of standard questions are given, which help to enhance knowledge of the topic. It has 14 direct questions aiming to have good practice of the sum of n terms formula; then there are six-word problems in which the sum of n terms formula is used trickly
NCERT solutions for exercise 5.3 Class 10 Maths chapter 5 Arithmetic progression discusses the sum of n terms of the Arithmetic Progression , which provides a wonderful practice of this topic by presenting the standard set of problems. 10th class Maths exercise 5.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.
Apply to Aakash iACST Scholarship Test 2024
Arithmetic Progressions Class 10 Chapter 5 Exercise: 5.3
Q1 (i) Find the sum of the following APs: to
terms.
Answer:
Given AP is
to
terms
Here,
And
Now, we know that
Therefore, the sum of AP to
terms is 245
Q1 (ii) Find the sum of the following APs: to
terms.
Answer:
Given AP is
to
terms.
Here,
And
Now, we know that
Therefore, the sum of AP to
terms. is -180
Q1 (iii) Find the sum of the following APs: to
terms.
Answer:
Given AP is
to
terms..
Here,
And
Now, we know that
Therefore, the sum of AP to
terms. is 5505
Q1 (iv) Find the sum of the following APs: to
terms.
Answer:
Given AP is
to
terms.
Here,
And
Now, we know that
Therefore, the sum of AP to
terms. is
Q2 (i) Find the sums given below :
Answer:
Given AP is
We first need to find the number of terms
Here,
And
Let suppose there are n terms in the AP
Now, we know that
Now, we know that
Therefore, the sum of AP is
Q2 (ii) Find the sums given below :
Answer:
Given AP is
We first need to find the number of terms
Here,
And
Let suppose there are n terms in the AP
Now, we know that
Now, we know that
Therefore, the sum of AP is 286
Q2 (iii) Find the sums given below :
Answer:
Given AP is
We first need to find the number of terms
Here,
And
Let suppose there are n terms in the AP
Now, we know that
Now, we know that
Therefore, the sum of AP is -8930
Q3 (i) In an AP: given ,
,
, find
and
.
Answer:
It is given that
Let suppose there are n terms in the AP
Now, we know that
Now, we know that
Therefore, the sum of the given AP is 440
Q3 (ii) In an AP: given ,
, find
and
.
Answer:
It is given that
Now, we know that
Therefore, the sum of given AP is 273
Q3 (iii) In an AP: given find
and
.
Answer:
It is given that
Now, we know that
Therefore, the sum of given AP is 246
Q3 (iv) In an AP: given find
and
Answer:
It is given that
Now, we know that
On solving equation (i) and (ii) we will get
Now,
Therefore, the value of d and 10th terms is -1 and 8 respectively
Q3 (vi) In an AP: given find
and
.
Answer:
It is given that
Now, we know that
n can not be negative so the only the value of n is 5
Now,
Therefore, value of n and nth term is 5 and 34 respectively
Q3 (vii) In an AP: given find
and
.
Answer:
It is given that
Now, we know that
Now, we know that
Now, put this value in (i) we will get
Therefore, value of n and d are respectively
Q3 (viii) In an AP: given find
and
.
Answer:
It is given that
Now, we know that
Now, we know that
Value of n cannot be negative so the only the value of n is 7
Now, put this value in (i) we will get
a = -8
Therefore, the value of n and a are 7 and -8 respectively
Q3 (ix) In an AP: given find
.
Answer:
It is given that
Now, we know that
Therefore, the value of d is 6
Q3 (x) In an AP: given and there are total
terms. Find
.
Answer:
It is given that
Now, we know that
Now, we know that
Therefore, the value of a is 4
Q4 How many terms of the AP: must be taken to give a sum of
?
Answer:
Given AP is
Here,
And
Now , we know that
Value of n can not be negative so the only the value of n is 12
Therefore, the sum of 12 terms of AP must be taken to give a sum of
.
Answer:
It is given that
Now, we know that
Now, we know that
Now, put this value in (i) we will get
Therefore, value of n and d are respectively
Answer:
It is given that
Now, we know that
Now, we know that
Therefore, there are 38 terms and their sun is 6973
Q7 Find the sum of first terms of an AP in which
and
nd term is
.
Answer:
It is given that
Now, we know that
Now, we know that
Therefore, there are 22 terms and their sum is 1661
Q8 Find the sum of first terms of an AP whose second and third terms are
and
respectively.
Answer:
It is given that
And
Now,
Now, we know that
Therefore, there are 51 terms and their sum is 5610
Q9 If the sum of first terms of an AP is
and that of
terms is
, find the sum of first
terms.
Answer:
It is given that
Now, we know that
Similarly,
On solving equation (i) and (ii) we will get
a = 1 and d = 2
Now, the sum of first n terms is
Therefore, the sum of n terms is
Q10 (i) Show that form an AP where an is defined as below :
Also find the sum of the first
terms.
Answer:
It is given that
We will check values of for different values of n
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to 7 and common difference (d) equals to 4
Now, we know that
Therefore, the sum of 15 terms is 525
Q10 (ii) Show that form an AP where an is defined as below :
. Also find the sum of the first
terms in each case.
Answer:
It is given that
We will check values of for different values of n
and so on.
From the above, we can clearly see that this is an AP with the first term(a) equals to 4 and common difference (d) equals to -5
Now, we know that
Therefore, the sum of 15 terms is -465
Answer:
It is given that
the sum of the first terms of an AP is
Now,
Now, first term is
Therefore, first term is 3
Similarly,
Therefore, sum of first two terms is 4
Now, we know that
Now,
Similarly,
Q12 Find the sum of the first positive integers divisible by
.
Answer:
Positive integers divisible by 6 are
6,12,18,...
This is an AP with
Now, we know that
Therefore, sum of the first positive integers divisible by
is 4920
Q13 Find the sum of the first multiples of
.
Answer:
First 15 multiples of 8 are
8,16,24,...
This is an AP with
Now, we know that
Therefore, sum of the first 15 multiple of 8 is 960
Q14 Find the sum of the odd numbers between and
.
Answer:
The odd number between 0 and 50 are
1,3,5,...49
This is an AP with
There are total 25 odd number between 0 and 50
Now, we know that
Therefore, sum of the odd numbers between and
625
Answer:
It is given that
Penalty for delay of completion beyond a certain date is Rs for the first day, Rs
for the second day, Rs
for the third day and penalty for each succeeding day being Rs
more than for the preceding day
We can clearly see that
200,250,300,..... is an AP with
Now, the penalty for 30 days is given by the expression
Therefore, the penalty for 30 days is 27750
Answer:
It is given that
Each price is decreased by 20 rupees,
Therefore, d = -20 and there are total 7 prizes so n = 7 and sum of prize money is Rs 700 so
Let a be the prize money given to the 1st student
Then,
Therefore, the prize given to the first student is Rs 160
Now,
Let is the prize money given to the next 6 students
then,
Therefore, prize money given to 1 to 7 student is 160,140,120,100,80,60.40
Answer:
First there are 12 classes and each class has 3 sections
Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1. Thus every class will plant 3 times the number of their class
Similarly,
No. of trees planted by 3 sections of class 1 = 3
No. of trees planted by 3 sections of class 2 = 6
No. of trees planted by 3 sections of class 3 = 9
No. of trees planted by 3 sections of class 4 = 12
Its clearly an AP with first term (a) = 3 and common difference (d) = 3 and total number of classes (n) = 12
Now, number of trees planted by 12 classes is given by
Therefore, number of trees planted by 12 classes is 234
[ Hint : Length of successive semicircles is with centres at
respectively.]
Answer:
From the above-given figure
Circumference of 1st semicircle
Similarly,
Circumference of 2nd semicircle
Circumference of 3rd semicircle
It is clear that this is an AP with
Now, sum of length of 13 such semicircles is given by
Therefore, sum of length of 13 such semicircles is 143 cm
Answer:
As the rows are going up, the no of logs are decreasing,
We can clearly see that 20, 19, 18, ..., is an AP.
and here
Let suppose 200 logs are arranged in 'n' rows,
Then,
Now,
case (i) n = 25
But number of rows can not be in negative numbers
Therefore, we will reject the value n = 25
case (ii) n = 16
Therefore, the number of rows in which 200 logs are arranged is equal to 5
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?
[ Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is ]
Answer:
Distance travelled by the competitor in picking and dropping 1st potato
Distance travelled by the competitor in picking and dropping 2nd potato
Distance travelled by the competitor in picking and dropping 3rd potato
and so on
we can clearly see that it is an AP with first term (a) = 10 and common difference (d) = 6
There are 10 potatoes in the line
Therefore, total distance travelled by the competitor in picking and dropping potatoes is
Therefore, the total distance travelled by the competitor in picking and dropping potatoes is 370 m
It's about the sum of n terms of the Arithmetic Progression, it contains a variety of standard questions to give the depth knowledge about the relevant topic. Firstly, it includes the questions in which the formula for the sum of n terms is applied directly; then, it has some slightly tricky questions in which it requires a good understanding of the formula for the sum of n terms. Exercise 5.3 Class 10 Maths -Arithmetic Progression is the progression in which the difference between two consecutive terms is constant. Using this very concept, the sum of n terms of the Arithmetic Progression could be calculated. The NCERT solutions for Class 10 Maths exercise 5.3 mainly focuses on the sum of n terms of the Arithmetic Progression which could be computed efficiently by having clarity regarding the basic concepts of the chapter and which are cleared in preceding exercises; this exercise takes a step further to the topic Arithmetic Progression Exercise 5.3 Class 10 Maths. Students can quickly go through the Arithmetic Progressions Class 10 Notes to revise all concepts all together.
Also see-
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
Any two successive terms in Arithmetic Progression differ by a constant numerical value.
It might be simply determined by using the fact that any two successive phrases differ by a constant numerical value. Merely add the difference (n-1) times to the first term of the Arithmetic Progression to compute it.
By sum of n terms of the Arithmetic Progression, it means the sum of all the terms of the Arithmetic Progression until the nth term.
Yes, it could be. It happens only when it has negative terms in the progression.
Yes, it might be in a fraction, but the Arithmetic Progression's number of terms cannot be in fraction.
Yes, it could be done conveniently by using the sum of n terms formula, and since the sum would be given already, we all have to put the values in the formula and calculate the value of n in the formula.
Simply differentiate the (n-1)th word from the nth term to discover the Arithmetic Progression's Common Difference.
According to this exercise, the sum of n terms of the Arithmetic Progression is the mathematical sum of all the terms of the Arithmetic Progression till the nth term.
The questions are based on the concept of the sum of n terms of ARithmetic Progression. To give a thorough practice on the topic, there are a set of standard questions available in the exercise; it contains direct formula-based questions and some word problems to enhance the understanding of the concept.
Late Fee Application Date:22 July,2024 - 31 July,2024
Late Fee Application Date:22 July,2024 - 31 July,2024
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.
Generally, passing 10th CBSE with a minimum percentage (often 50%) is the basic eligibility for diploma courses. Some institutes might have specific subject requirements depending on the diploma specialization.
There is a wide range of diploma courses available in various fields like engineering (e.g., mechanical, civil, computer science), computer applications, animation, fashion design, hospitality management, and many more.
You can pursue diplomas at various institutions like:
Register for Tallentex '25 - One of The Biggest Talent Encouragement Exam
Get up to 90% scholarship on NEET, JEE & Foundation courses
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
Accepted by more than 11,000 universities in over 150 countries worldwide
Register now for PTE & Save 5% on English Proficiency Tests with ApplyShop Gift Cards