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Class 10 (SSC Board)Mathematics & Statistics2026-27 Syllabus

Chapter 3 Arithmetic Progression Practice Set 3.4 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 3 Arithmetic Progression Practice Set 3.4. Step-by-step solved exercises, numerical problems, and digest answers.

6 Solved Questions5 Diagrams751 words

Practice Set 3.4 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

Question 1 Maharashtra Board Solution
On 1st Jan 2016, Sanika decides to save ₹ 10, ₹ 11 on second day, ₹ 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?
Solution & Step-by-Step Answer:
∴ Sanika’s total saving on 31st December 2016 would be ₹ 70455.

Question 2 Maharashtra Board Solution
A man borrows ₹ 8000 and agrees to repay with a total interest of ₹ 1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 40. Find the amount of the first and last instalment.
Solution & Step-by-Step Answer:
i. The instalments are in A.P. Amount repaid in 12 instalments (S12) = Amount borrowed + total interest = 8000 + 1360 ∴ S12 = 9360 Number of instalments (n) = 12 Each instalment is less than the preceding one by ₹ 40. ∴ d = -40 ∴ Amount of the first instalment is ₹ 1000 and that of the last instalment is ₹ 560.

Question 3 Maharashtra Board Solution
Sachin invested in a national saving certificate scheme. In the first year he invested ₹ 5000, in the second year ₹ 7000, in the third year ₹ 9000 and so on. Find the total amount that he invested in 12 years.
Solution & Step-by-Step Answer:
i. Amount invested by Sachin in each year are as follows: 5000, 7000, 9000, … The above sequence is an A.P. ∴ a = 5000, d = 7000 – 5000 = 2000, n = 12


∴ The total amount invested by Sachin in 12 years is ₹ 1,92,000.

Question 4 Maharashtra Board Solution
There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium?
Solution & Step-by-Step Answer:
i. The number of seats arranged row-wise are as follows: 20, 22, 24, The above sequence is an A.P. ∴ a = 20, d = 22 – 20 = 2, n = 27

ii. tn= a + (n – 1)d
∴ t15= 20 + (15 – 1)2
= 20 + 14 × 2
= 20 + 28
∴ t15= 48
∴ The number of seats in the 15throw is 48.

∴ Total seats in the auditorium are 1242.

Question 5 Maharashtra Board Solution
Kargil’s temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was 5°C more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was -30° Celsius then find the temperature on the other five days.
Solution & Step-by-Step Answer:
Let the temperatures from Monday to Saturday in A.P. be a, a + d, a + 2d, a + 3d, a + 4d, a + 5d. According to the first condition, (a) + (a + 5d) = (a + d) + (a + 5d) + 5° ∴ d = -5° According to the second condition, a + 2d = -30° ∴ a + 2(-5°) = -30° ∴ a – 10° = -30° ∴ a = -30° + 10° = -20° ∴ a + d = -20° – 5° = – 25° a + 3d = -20° + 3(- 5°) = -20° – 15° = -35° a + 4d = -20° + 4(-5°) = -20° – 20° = -40° a + 5d = -20° + 5(-5°) = -20° – 25° = -45° ∴ The temperatures on the other five days are -20°C, -25° C, -35° C, -40° C and -45° C.
Question 6 Maharashtra Board Solution
On the world environment day tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.
Solution & Step-by-Step Answer:
i. The number of frees planted row-wise are as follows: 1,2,3,… The above sequence is an A.P. ∴ a = 1, d = 2 – 1 = 1,n = 25 ∴ The total number of trees in 25 rows are 325.