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

Here you get the CBSE Class 10 Mathematics chapter 5, Arithmetic Progressions: NCERT Exemplar Problems and Solutions (Part-IIID). This part of the chapter includes solutions of Question Number 30 to 35 from Exercise 5.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Arithmetic Progressions. This exercise comprises of only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

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

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.

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

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

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

Exercise 5.3

Short Answer Type Questions (Q. No. 30-35):

Question. 30 Find the sum of last ten terms of the AP 8, 10, 12, ....., 126.

Solution.

For finding, the sum of last ten terms, we write the given AP in reverse order.

Question. 31 Find the sum of first seven numbers which are multiples of 2 as well as of 9.

Solution.

To find, the list of  numbers which are multiples of 2 as well as of 9, we first take the LCM of 2 and 9 which is 18.

So, the series of numbers which are multiples of 2 as well as of 9 is: 18, 36, 54,....

Here, first term, a = 18

And, common difference, d = 36 – 18 = 18

Question. 32 How many terms of the AP −15, −13, −11, ..... are needed to make the sum −55?

Solution.

Given AP is −15, −13, −11, .....

Let n number of terms this of AP are needed to make the sum −55.

i.e., S_n = −55

Here, first term, a = −15

And, common difference, d = −13 – (−15) = −13 + 15 = 2

Hence, either 5 or 11 terms are needed to make the sum −55.

Question. 33 The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is -30 and the common difference is 8. Find n.

Solution.

Question. 34 Kanika was given her pocket money on Jan 1^st, 2008. She puts Rs. 1 on day 1, Rs. 2 on day 2, Rs. 3 on day 3 and continued doing so till the end of the month, from this money into her piggy bank she also spent Rs. 204 of her pocket money, and found that at the end of the month she still had Rs. 100 with her. How much was her pocket money for the month.

Solution.

Let her pocket money be Rs. x.

Out of Rs. x, money put in piggy bank from Jan. 1 to Jan. 31 = 1 + 2 + 3 + 4 + … + 31.

Which forms an AP in first term, common difference and number of terms being 1, 1 and 31 respectively.

Sum of first 31 terms is given as:

Question. 35 Yasmeen saves Rs. 32 during the first month, Rs. 36 in the second month and Rs. 40 in the third month. If she continues to save in this manner, in how many moths will she save Rs. 2000?

Solution.

Amount saved during the first month = Rs. 32

Amount saved during the second month = Rs. 36

Amount saved during the third month = Rs. 40

Thus we have an arithmetic progression 32, 36, 40, ......

Here, first term, a = 32,

And, common difference, d = 36 – 32 = 4

Let Rs. 2000 will be saved during n months.

i.e., S_n = Rs. 2000

Ignore n = −40 as a month cannot be negative.

Hence, Rs. 2000 will be saved in 25 months.                        

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