Maharashtra Board Class 10 Math Part 1 Solution Chapter 3 Practice Set 3.3 – 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 |
3.3 |
Practice Set 3.3
(1) First term and common difference of an A.P. are 6 and 3 respectively; find S27.
a = 6, d = 3, s27 =?
Sn = n/2 [ + (n-1) d]
S27 = 27/2 [12 + (27 – 1) ]
= 27/2 ×
= 27 × 45 =
Solution:
Given, a = 6, d = 3, s27 =?
sn = n/2 [2a + (n-1) d]
= 27/2 [2×6 + (27-1) ×3]
= 27/2 × (12 + 78)
= 27/2 × 90
= 1215
(2) Find the sum of first 123 even natural numbers.
Solution:
Given, n = 123. for even natural numbers the AP is
2, 4, 6, 8, …… n123
Now, a = 2; t2 – t1 = 4 – 2 = 2
∴ d = 2 t3 – t2 = 6 – 4 = 2
∴ sn = n/2 [2a + (n – 1) d]
Or, s123 = 123/2 [2×2 + (123 – 1) × 2]
= 123/2 × (4 + 244)
= 123/2 × 248
= 15252.
∴ Sum of first 123 natural numbers is 15252.
(3) Find the sum of all even numbers between 1 and 350.
Solution:
The AP for all even numbers from 1 to 350 is
2, 4, 6, 8, …. 350
∴ t1 = 2, tn = 350
∴ a = 2; t2 – t1 = 4 – 2 = 2
d = 2 t3 – t2 = 6 – 4 = 2
∴ For tn = 350
tn = a + (n – 1) d
Or, 350 = 2 + (n – 1) 2
Or, 350 = 2 + 2n – 2
Or, n = 350/2 = 175
∴ sn = n/2 [t1 + tn]
Or, s175 = 175/2 [2 + 350]
= 175/2 × 352
= 175 × 176
= 30800
∴ The sum of even numbers between 1 and 350 is 30800.
(4) In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.
Solution:
Given, t19 = 52; t36 = 128; n = 56
∴ t19 = a + (n – 1) d
Or, t19 = a + (19 – 1) d
Or, 52 = a + 18d
Or, a + 18d = 52 —– (i)
t38 = a + (n – 1) d
Or, 128 = a + (38 – 1) d
Or, a + 37d = 128 —– (ii)
(5) Complete the following activity to find the sum of natural numbers between 1 and 140 which are divisible by 4.
Solution:
Between 1 and 140, natural numbers divisible by 4
4, 8, ….., 136
a = 4, t2 – t1 = 8 – 4 = 4
∴ d = 4
∴ tn = 136 given,
∴ tn = a + (n – 1) d
Or, 136 = 4 + (n – 1) × 4
Or, 136 = 4 + 4n – 4
Or, n = 136/4 = 34
∴ n = 34, a = 4, d = 4
Now, sn = n/2 [2a + (n-1) d
Or, s34 = 34/2 [2×4 + (34 – 1)4]
= 17 [8 + 132]
= 17 × 140
= 2380
∴ Sum of numbers from 1 to 140 divisible by 4 = 2380.
(6) Sum of first 55 terms in an A.P. is 3300, find its 28th term.
Solution:
Given, S55 = 3300
We know, Sn = n/2 [2a + (n – 1) d]
Or, 3300 = 55/2 [24 + (55 – 1) d]
Or, 3300 = 55/2 × (2a + 54d)
Or, 3300×2/55 = 2a + 54d
Or, 3300×2/55 = (a + 27d)
Or, a + 27d = 60 —- (i)
Now, we know tn = a + (n – 1) d [Now n = 28]
Or, t28 = a + (28 – 1) d
Or, t28 = a + 27d
Now, substituting the value 7a + 27 d from equation (i)
Or, t28 = 60
∴ The 28th term 7th AP is 60.
(7) In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a – d, a, a + d.)
Solution:
Given, a – d + a + + a + d = 27 [When ‘a’ is the first from and d is the common difference]
or, 3a = 27
Or, a = 9
Now, (ATQ) (a – d) × a × (a + d) = 504
Or, (9 – 6) × 9 (9 + d) = 504
Or, (9 – d) (9 + d) = 504/4
Or, 92 – d2 = 56 [∵ a2 – b2 = (a + b) (a – b)
Or, d2 = 81 – 56
Or, d = √25
= 5
∴ The three terms are a – d = 9 – 5 = 4,
a = 9;
a + d = 9 + 5 = 14
(8) Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.
(Assume the four consecutive terms in A.P. are a – d , a, a + d, a + 2d.)
Solution:
Let, the tour terms of AP be
a, a + d, a + 2d, a + 3d when, a is the first term and d is the common difference
Now, (ATQ)
a + a + d + a + 2d + a + 3d = 12
Or, 4a + 6d = 12
Or, 2a + 3d = 12/2
Or, given, t3 + t4 = 14
We know, tn = a + (n – 1) d
a, t3 = a + (3 – 1) d
= a + 2d
t4 = a + (4-1) d
= a + 3d
Nov, t3 + t4 = 14
Or, a + 2d + a + 3d = 14
Or, 2a + 5d = 14 —– (ii)
Subtracting equation (i) from equation (ii) we get
Nov, substituting the value of ‘d’ is equation (i)
2a + 3d = 6
Or, 2a + 3×4 = 6
Or, 2a = 6 – 12
Or, a = -6/2 = -3
∴ 1st term = a = -3
2nd term = a + d = – 3 + 4 = 1
3rd term = a + 2d = – 3 + 4 × 2 = – 3 + 8 = 5
4th term = a + 3d = – 3 + 3 × 4 = – 3 + 12 = 9
∴ The four terms are -3, 1, 5 and 9.
(9) If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.
Solution:
Let, a be the first term.
d be the common difference
(ATQ)
t9 = 0
We know, tn = a + (n – 1) d
Or, t9 = a + (9 – 1) d
Or, 0 = a + 8d =
Or, a + 8d = 0
Or, a = -8d —- (i)
Similarly t29 = a + (29 – 1) d
= a + 28d
t19 = a + (19 – 1) d
= a + 18d
Putting the value of a from equation (i) we get.
t29 = a + 28d
= – 8d + 28d
= 20d
t19 = a + 18d
= -8d +18d
= 10d
Now, t29/t19 = 20d/10d
Or, t29 = 2t19
∴ The 29th term is twice the 19th term (Hence Proved)