Maharashtra Board Class 10 Maths Chapter 3 Arithmetic Progression Set 3.4 Solutions

Question 1. On 1st Jan 2016, Sanika decides to save Rs. 10, Rs. 11 on second day, Rs. 12 on third day. If she decides to save like this, then on 31st Dec 2016 what would be her total saving?
Answer:
(i) Sanika's daily savings of 2016 are as follows:
10, 11, 12, ......
The above sequence is an A.P.
.. \(a = 10\), \(d = 11 - 10 = 1\),
\(n = 366\) (2016 is a leap year)
(ii) \(S_n = \frac{n}{2} [2a + (n-1)d]\)
.. \(S_{366} = \frac{366}{2} [2 (10) + (366-1) 1]\)
\( = 183 (20 + 365 \times 1)\)
\( = 183 (20+365)\)
\( = 183 \times 385\)
.. \(S_{366} = 70455\)
.. Sanika's total saving on 31st December 2016 would be Rs. 70455.
In simple words: Sanika's daily savings form an arithmetic progression. Since 2016 was a leap year, there are 366 days. We calculated the sum of this A.P. to find her total savings.

🎯 Exam Tip: Remember to correctly identify if the year is a leap year when dealing with daily savings problems, as it affects the value of 'n'.

Question 2. A man borrows Rs. 8000 and agrees to repay with a total interest of Rs. 1360 in 12 monthly instalments. Each instalment being less than the preceding one by 40. Find the amount of the first and last instalment.
Answer:
(i) The instalments are in A.P.
Amount repaid in 12 instalments (\(S_{12}\))
= Amount borrowed + total interest
= \(8000 + 1360\)
.. \(S_{12} = 9360\)
Number of instalments (\(n\)) = 12
Each instalment is less than the preceding one by 40.
.. \(d = -40\)
(ii) \(S_n = \frac{n}{2} [2a + (n-1)d]\)
.. \(S_{12} = \frac{12}{2} [2a + (12-1) (-40)]\)
\(9360 = 6[2a + (11) (-40)]\)
\(9360 = 6(2a - 440)\)
.. \(\frac{9360}{6} = 2a - 440\)
\(1560 = 2a - 440\)
.. \(1560 + 440 = 2a\)
\(2000 = 2a\)
.. \(a = \frac{2000}{2}\)
.. \(a = 1000\)
(iii) \(t_n = a + (n-1) d\)
\(t_{12} = 1000 + (12-1) (-40)\)
\( = 1000 + 11 (-40)\)
\( = 1000 - 440\)
.. \(t_{12} = 560\)
.. Amount of the first instalment is Rs. 1000 and that of the last instalment is Rs. 560.
In simple words: This problem involves an arithmetic progression where the installments decrease over time. We used the sum formula \(S_n\) to find the first installment ('a') and then the nth term formula \(t_n\) to find the last installment.

🎯 Exam Tip: Be careful with the common difference 'd' when installments are decreasing; it will be a negative value.

Question 3. Sachin invested in a national saving certificate scheme. In the first year he invested Rs. 5000, in the second year Rs. 7000, in the third year Rs. 9000 and so on. Find the total amount that he invested in 12 years.
Answer:
(i) Amount invested by Sachin in each year are as follows:
5000, 7000, 9000, ...
The above sequence is an A.P.
.. \(a = 5000\), \(d = 7000 - 5000 = 2000\), \(n = 12\)
(ii) \(S_n = \frac{n}{2} [2a + (n-1)d]\)
.. \(S_{12} = \frac{12}{2} [2 (5000) + (12 - 1) 2000]\)
\( = 6 (10000 + 11 \times 2000)\)
\( = 6 (10000 + 22000)\)
\( = 6 (32000)\)
.. \(S_{12} = 192000\)
.. The total amount invested by Sachin in 12 years is Rs. 1,92,000.
In simple words: Sachin's yearly investments form an arithmetic progression. To find the total amount invested over 12 years, we calculated the sum of this A.P.

🎯 Exam Tip: When calculating total investments over several years, ensure you correctly identify the first term, common difference, and the number of years 'n'.

Question 4. 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?
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) \(t_n = a + (n - 1)d\)
.. \(t_{15} = 20 + (15 - 1)2\)
\( = 20 + 14 \times 2\)
\( = 20 + 28\)
.. \(t_{15} = 48\)
.. The number of seats in the 15th row is 48.
(iii) \(S_n = \frac{n}{2} [2a + (n-1)d]\)
.. \(S_{27} = \frac{27}{2} [2 (20) + (27 - 1) 2]\)
\( = \frac{27}{2} (40 + 26 \times 2)\)
\( = \frac{27}{2} (40 + 52)\)
\( = \frac{27}{2} \times 92\)
\( = 27 \times 46\)
.. \(S_{27} = 1242\)
.. Total seats in the auditorium are 1242.
In simple words: The number of seats per row forms an arithmetic progression. We used the term formula \(t_n\) to find seats in a specific row and the sum formula \(S_n\) to find the total seats in the auditorium.

🎯 Exam Tip: This problem requires calculating both a specific term (\(t_n\)) and the sum of the series (\(S_n\)); make sure to apply the correct formula for each part.

Question 5. 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.
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.
In simple words: We used the properties of an arithmetic progression and the given conditions to form equations, solving for 'a' (Monday's temperature) and 'd' (common difference) to find all daily temperatures.

🎯 Exam Tip: Clearly define the terms of the A.P. (e.g., temperatures for each day) and meticulously set up equations based on the given conditions to avoid errors.

Question 6. 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.
Answer:
(i) The number of trees planted row-wise are as follows:
1,2,3,...
The above sequence is an A.P.
.. \(a = 1\), \(d = 2 - 1 = 1\), \(n = 25\)
(ii) \(S_n = \frac{n}{2} [2a + (n-1)d]\)
.. \(S_{25} = \frac{25}{2} [2 (1) + (25 - 1) 1]\)
\( = \frac{25}{2} (2+24)\)
\( = \frac{25}{2} \times 26\)
\( = 25 \times 13 = 325\)
.. The total number of trees in 25 rows are 325.
In simple words: The number of trees in each row forms an arithmetic progression (1, 2, 3...). To find the total trees in 25 rows, we calculate the sum of this A.P.

🎯 Exam Tip: Recognize simple number sequences like natural numbers (1, 2, 3...) as arithmetic progressions to easily apply the sum formula.