Maharashtra Board Class 10 Math Part 1 Solution Chapter 3 Practice Set 3.4 – Arithmetic Progression
Balbharati Maharashtra Board Class 10 Math Part 1 Solution Chapter 3: Arithmetic Progression. Marathi or English Medium Students of Class 10 get here Arithmetic Progression full Practice set Solution.
|
Std |
Maharashtra Class 10 |
| Subject |
Math Part 1 Solution |
|
Chapter |
Arithmetic Progression |
| Practice set |
3.4 |
Practice Set 3.4
(1) 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:
Savings of Sanika is in an A.P as follows –
₹10, ₹17, ₹12, …..
∴ For 1st December she started he savings.
∴ (ATQ) d = t2 – t1
= 12 – 11
= 1
Now, Here no. of terms is from 1st June to 31st December.
∴ n = 366 [∵ 2016 is a leap year 366 days total]
a = 10; d = t2 – t1
= 11 – 10 = 1
∴ Sn = n/2 [2a + (n – 1) d]
= 366/2 × [2×10 + (366 – 1) i]
= 362/2 × [2 × 10 + 365]
= 183 × 385
= RS 70,455
∴ Sanika saves RS 70455 at the end of the year.
(2) 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:
Money borrowed by the men = RS 8000
Total interest = RS 1360
Total money to be paid after 10 months = 8000 + 1360
= RS 9360
Now, (ATQ) each installment is RS 40 less than the previous one.
∴ d = -40
n = 12
S12 = 9360
Now, sn = n/2 [2a + (n – 1) d]
Or, s12 = 12/2 [2a + (12 – 1) -40]
Or, 9360 = 6 (2a – 440)
Or, 1560 = 2a – 440
Or, 2a = 1560 + 440
Or, a = 2000/2 = 1000
∴ Now, a12 = a + (12 – 1)
= 1000 + 11 x – 40
= 560
∴ The first installment is RS 1000 and the last one is RS 560.
(3) 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:
Investments each years is as follows.
5000, 7000, 9000 …..
∴ d = t2 – t1
a = 5000
n = 12 years
= 7000 – 5000
= 2000
∴ Sn = 12/2 [2a + (n – 1) d]
Or, s12 = 12/2 [2 × 5000 + (12 – 1) d]
Or, s12 = 6 [10000 + 22000]
= 6 × 32000
= 192000
∴ Total investment is 12 years = RS 192000
(4) 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?
Solution:
(ATQ)
Total no. of rows = 2t
The A.P of seats are as follows, 20, 22, 24, —-
∴ a = 20
n = 27
d = t2 – t1 = 22 – 20
= 2
a15 = a + (15 – 1) d
= 20 + 14 × 2
= 20 + 28
= 48
∴ There are 48 seats in row 15th
Now, total no. of seats in auditorium
= Sn = S27
∴ Sn = n/2 [2 + (n-1) d]
Or, S27 = 27/2 [2×20 + (27 – 1) ×2]
= 27/2 × (40 + 26 × 2)
= 27/2 × 92 = 1242
∴ There are 1242 no. of seats in the auditorium.
(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.
Solution:
Temps are in A.P.
Now let Monday temp be M
Let Saturdays temp be S
Let Tuesdays temp be T
∴ (ATQ) M + S = T + S + 5
Or, M – T = 5
∴ Or, t1 – t2 = 5
Or, t2 – t = -5
∴ d = -5
∴ Temp of Tuesday = Wednesday – d
= – 30 – (-5)
= – 30 + 5
= – 26°
Temp of Monday = Tuesday – d
= – 25 – (-5)
= – 25 + 5
= -20°
Temp of Thursday = Wednesday + d
= -30 + (-S)
= – 35°
Temp of Friday = Thursday + d
= – 35 – 5
= 40°
Temp of Saturday = Friday + d
= – 40 – 5 = – 45°
(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.
Solution:
Given, the trees are in A.P as follows 1, 2, 3, ….
Total rows = 25
∴ n = 25
a = 1
d = t2 – t1
= 2 – 1
= 1
∴ S15 = n/2 [2a + (n – 1) d]
= 25/2 [2×1 + (25 – 1) × 1]
= 25/2 (2 + 24)
= 25/2 × 26
= 325
∴ There are total 325 trees.
