NCERT Exemplar Solution for CBSE Class 10 Mathematics: Arithmetic Progressions (Part-IVB)

Jul 10, 2017 17:39 IST

Class 10 Maths NCERT Exemplar, Arithmetic Progressions NCERT Exemplar Problems, NCERT Exemplar Problems, Class 10 Chapter 5 NCERT ExemplarHere you get the CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar Problems and Solutions (Part-IVB). This part of the chapter includes solutions of Question Number 6 to 10 from Exercise 5.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Arithmetic Progressions. This exercise comprises of only the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

CBSE Class 10 Mathematics Syllabus 2017-2018

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Arithmetic Progressions:

Exercise 5.4

Long Answer Type Questions (Q. No. 6-10):

Question. 6 The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term and also the ratio of the sum of the first five terms to the sum of the first 21 terms.


Let a and d be the first term and common difference of an AP.

Now nth term of an AP is given as:

            an = a + (n – 1)d

∴ 11th term of AP, a11 = a + 10d

And, 18th term of AP, a18 = a + 17d

According to question,


So, ratio of the sum of the first five terms of the AP to the sum of the first 21 terms is given as

            S5 : S21 = 30d  : 294d = 5 : 49

Question. 7 Show that the sum of an AP whose first term is a, the second term b and the last

Hence proved.

Question. 8 Solve the equation – 4 + (–1) + 2 + ... + x = 437.


Given equation is, – 4 – 1 + 2 + .... + x = 437                                    .... (i)

LHS of equation (i) forms an AP with

First term, a = – 4,

Common difference, d = 3,

And last term, l = an  = x


Here, x cannot be negative as for x = −53, n will be negative which is not possible

Hence, the required value of x is 50.

Question. 9 Jaspal singh repays his total loan of Rs. 118000 by paying every month starting with the first installment of Rs. 1000. If he increases the installment by Rs. 100 every month, what amount will be paid by him in the 30th installment? What amount of loan does he still have to pay after the 30th installment?


Jaspal singh takes total loan = Rs. 118000

Monthly instalments paid by Jaspal are Rs. 1000, Rs. 1100, Rs. 1200, …. Which form an AP with:

First term, a = 1000

Common difference, d = 100

30th term of this AP, a30 = 1000 + (30 − 1)100           [Using, an = a + (n−1)d]

                                    = 1000 + 29 × 100

                                    = 1000 + 2900 = 3900

So, Rs. 3900 will be paid by Jaspal in the 30th installment.

Now, Total amount paid in 30 installments is = 1000 + 1100 + 1200 + ..... + 3900

Here, First term, a = 1000,

Last term, l = 3900

And, number of terms, n = 30

Total amount that Jaspal still have to pay after the 30th installment

            = (Amount of loan) – (Sum of 30 installments)

            = 118000 – 73500 = Rs. 44500

Question. 10 The students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags.
Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance she did cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?


Thus Ruchi have to place 13 flags on the left of 14th flag and 13 flags on its right.

Similarly, for placing third flag and return his initial position, distance travelled = 4+ 4= 8m

For placing fourth flag and return his initial position, distance travelled = 6+ 6 = 12 m.

For placing fourteenth flag and return his initial position, distance travelled

                                                = 26 + 26 = 52m

Proceed same manner into her right position from middle most flag i.e., 14th flag.

Distance between two adjacent flags = 2m

Therefore distance travelled to place first flag = 4m

Distance travelled to place second flag = 8m

Distance travelled to place 13th flag = 52 m

Also, when Ruchi placed the last flag she returns his middle most position and collects her books. This distance also included in placed the last flag.

So, these distances form a series.

            4 + 8+ 12 + 16 + .... + 52        [For left side]

And     4 + 8 + 12 + 16 + ..... + 52      [For right side]

Here, first term, a = 4

Common difference, d = 8 – 4 = 4

And, number of terms, n = 13

Now, the maximum distance she travelled carrying a flag = Distance travelled by Ruchi during placing the 14th flag in her left position or 27th flag in her right position

            = (2 + 2 + 2 + ... + 13 times)

            = 2 × 13 = 26 m

Hence, the maximum distance she travelled carrying a flag is 26 m.

You may also like to read:

CBSE Class 10 NCERT Textbooks & NCERT Solutions

NCERT Solutions for CBSE Class 10 Maths

NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters

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