Exercise 2.5 - Detailed Solutions
1. Check whether the following sequences are in A.P.
(i) a - 3, a - 5, a - 7, ...
t2 - t1 = (a - 5) - (a - 3) = -2
t3 - t2 = (a - 7) - (a - 5) = -2
Since the common difference is constant (-2), it IS an A.P.
(ii) 1/2, 1/3, 1/4, 1/5, ...
t2 - t1 = 1/3 - 1/2 = -1/6
t3 - t2 = 1/4 - 1/3 = -1/12
Since t2 - t1 ≠ t3 - t2, it IS NOT an A.P.
(iii) 9, 13, 17, 21, 25, ...
13 - 9 = 4, 17 - 13 = 4, 21 - 17 = 4
Since the common difference is constant (4), it IS an A.P.
(iv) -1/3, 0, 1/3, 2/3, ...
0 - (-1/3) = 1/3, 1/3 - 0 = 1/3, 2/3 - 1/3 = 1/3
Since the common difference is constant (1/3), it IS an A.P.
(v) 1, -1, 1, -1, 1, ...
t2 - t1 = -1 - 1 = -2
t3 - t2 = 1 - (-1) = 2
Since t2 - t1 ≠ t3 - t2, it IS NOT an A.P.
2. First term 'a' and common difference 'd' are given below. Find the corresponding A.P.
Formula: A.P. = a, a + d, a + 2d, a + 3d, ...
(i) a = 5, d = 6
Terms: 5, 5 + 6, 5 + 2(6), 5 + 3(6)...
A.P.: 5, 11, 17, 23, ...
(ii) a = 7, d = -5
Terms: 7, 7 + (-5), 7 + 2(-5), 7 + 3(-5)...
A.P.: 7, 2, -3, -8, ...
(iii) a = 3/4, d = 1/2
t1 = 3/4
t2 = 3/4 + 1/2 = 5/4
t3 = 5/4 + 1/2 = 7/4
A.P.: 3/4, 5/4, 7/4, 9/4, ...
3. Find the first term and common difference of the A.P.s whose nth terms are given below.
(i) tn = -3 + 2n
For n = 1: a = t1 = -3 + 2(1) = -1
For n = 2: t2 = -3 + 2(2) = 1
Common difference d = t2 - t1 = 1 - (-1) = 2
First term (a) = -1, Common Difference (d) = 2
(ii) tn = 4 - 7n
For n = 1: a = t1 = 4 - 7(1) = -3
For n = 2: t2 = 4 - 7(2) = -10
Common difference d = t2 - t1 = -10 - (-3) = -7
First term (a) = -3, Common Difference (d) = -7
4. Find the 19th term of an A.P. -11, -15, -19, ...
Here, a = -11, d = -15 - (-11) = -4, n = 19
Formula: tn = a + (n - 1)d
t19 = -11 + (19 - 1)(-4)
t19 = -11 + 18(-4) = -11 - 72 = -83
19th term = -83
5. Which term of an A.P. 16, 11, 6, 1, ... is -54?
a = 16, d = 11 - 16 = -5, l = -54
Formula: n = [(l - a) / d] + 1
n = [(-54 - 16) / -5] + 1
n = [-70 / -5] + 1 = 14 + 1 = 15
-54 is the 15th term
6. Find the middle term(s) of an A.P. 9, 15, 21, 27, ..., 183.
a = 9, d = 15 - 9 = 6, l = 183
Total terms, n = [(183 - 9) / 6] + 1 = (174 / 6) + 1 = 29 + 1 = 30
Since n = 30 (even), middle terms are (n/2) and (n/2 + 1) → 15th and 16th terms.
t15 = 9 + (15 - 1)6 = 9 + 14(6) = 9 + 84 = 93
t16 = 93 + 6 = 99
Middle terms are 93 and 99
7. If 9 times ninth term is equal to 15 times fifteenth term, show that 6 times twenty-fourth term is zero.
Given: 9 × t9 = 15 × t15
9(a + 8d) = 15(a + 14d)
9a + 72d = 15a + 210d
15a - 9a + 210d - 72d = 0 ⇒ 6a + 138d = 0
Divide by 6: a + 23d = 0 ⇒ t24 = 0
Therefore, 6 × t24 = 6 × 0 = 0
Hence proved.
8. If 3 + k, 18 - k, 5k + 1 are in A.P., then find k.
Condition for A.P.: 2b = a + c
2(18 - k) = (3 + k) + (5k + 1)
36 - 2k = 6k + 4
36 - 4 = 6k + 2k ⇒ 8k = 32
k = 4
9. Find x, y, and z given that x, 10, y, 24, z are in A.P.
Here t2 = 10 and t4 = 24
t4 - t2 = 2d ⇒ 24 - 10 = 2d ⇒ 2d = 14 ⇒ d = 7
x = t1 = 10 - d = 10 - 7 = 3
y = t3 = 10 + d = 10 + 7 = 17
z = t5 = 24 + d = 24 + 7 = 31
x = 3, y = 17, z = 31
10. In a theatre, 20 seats in front row, 30 rows total. Each row has 2 additional seats than its front row. How many seats in last row?
a = 20, d = 2, n = 30
t30 = a + (30 - 1)d = 20 + 29(2) = 20 + 58 = 78
Seats in last row = 78
11. Sum of 3 consecutive terms in A.P. is 27 and product is 288. Find the terms.
Let terms be (a - d), a, (a + d)
Sum: (a - d) + a + (a + d) = 27 ⇒ 3a = 27 ⇒ a = 9
Product: (9 - d)(9)(9 + d) = 288
(81 - d2) = 288 / 9 = 32
d2 = 81 - 32 = 49 ⇒ d = ±7
If d = 7: Terms are 2, 9, 16
If d = -7: Terms are 16, 9, 2
Three terms are 2, 9, 16 (or 16, 9, 2)
12. Ratio of 6th and 8th term is 7:9. Find ratio of 9th term to 13th term.
Given: t6 / t8 = 7 / 9 ⇒ (a + 5d) / (a + 7d) = 7 / 9
9(a + 5d) = 7(a + 7d) ⇒ 9a + 45d = 7a + 49d ⇒ 2a = 4d ⇒ a = 2d
Required Ratio: t9 / t13 = (a + 8d) / (a + 12d)
Substitute a = 2d: (2d + 8d) / (2d + 12d) = 10d / 14d = 5 / 7
Ratio = 5:7
13. Temperature of Ooty Mon to Fri in A.P. Sum Mon to Wed is 0°C. Sum Wed to Fri is 18°C. Find temperatures.
Let temperatures be a - 2d, a - d, a, a + d, a + 2d
Mon to Wed Sum: (a - 2d) + (a - d) + a = 0 ⇒ 3a - 3d = 0 ⇒ a = d
Wed to Fri Sum: a + (a + d) + (a + 2d) = 18 ⇒ 3a + 3d = 18
Substitute d = a: 3a + 3a = 18 ⇒ 6a = 18 ⇒ a = 3°C, d = 3°C
Mon = 3 - 6 = -3°C | Tue = 3 - 3 = 0°C | Wed = 3°C | Thu = 6°C | Fri = 9°C
Temperatures: -3°C, 0°C, 3°C, 6°C, 9°C
14. Priya salary ₹15000, increases ₹1500/yr. Expenses ₹13000, increases ₹900/yr. How long to save ₹20000/month?
1st year savings = Salary - Expenses = 15000 - 13000 = ₹2000
Yearly increase in savings = Salary increase - Expense increase = 1500 - 900 = ₹600
Here, a = 2000, d = 600, target tn = 20000
20000 = 2000 + (n - 1)600
18000 = (n - 1)600 ⇒ n - 1 = 30 ⇒ n = 31
Time taken = 31 years
No comments:
Post a Comment