Arithmetic Progressions Test - 1

Here, a = first term, d = common difference, S_n = Sum of n terms, n = number of terms, t_n = n^th term
Since 17^th term = 0
a + (17 – 1)d = 0
a + 16d = 0 --- (1)

31^st term = a + (31 – 1)d
= a + 30d
= a + 16d + 14d
From equation (1),
a₃₁ = 0 + 14d
a₃₁ = 14d
This means 31^st term = 14d --- (2)

24^th term = a + (24 - 1)d
= a + 23d
= a + 16d + 7d
From equation (1),
a₂₄ = 0 + 7d
This means 24^th term = 7d --- (3)
From equations (2) and (3) we can say that the 31^st term is double of the 24^th term.

Suppose, first term = a, and common difference = d = -2
Since first term = 9 times 9^th term;
a = 9(a + 8d)
a = 9(a + 8 × (-2))
a = 9(a - 16)
a = 9a - 144
a - 9a = (-144)
-8a = (-144)
a = 18
So, 6^th term = 18 + (5 × (-2))
= 18 - 10
= 8

Let the consecutive terms be (a - d), a, (a + d).
Sum of these terms = 81
a - d + a + a + d = 81
3a = 81
a = 27, which means the numbers would become 27 - d, 27, 27 + d.
So, their product = (27 - d) × 27 × (27 + d) = 19,440
27² - d² = 19,440 ÷ 27
729 - d² = 720
-d² = 720 - 729
d² = 9
d= ±3
Let d = 3
The terms are:
27 - 3 = 24; 27; 27 + 3 = 30

Here, n = number of terms, a = first term, d = common difference, S_n = sum of n terms, t_n = n^th term
Sum of 10 terms = n/2 (2a + (n – 1)d)
14,520 = 10/2 (2a + (10 - 1)d)

14,520 = 5(2a + 9d)

2904 = 2a + 9d --- (1)

Sum of 20 terms = n/2 (2a + (n - 1)d)
2,68,100 = 20/2 (2a + (20 - 1) d)
26,810 = 2a + 19d --- (2)

Subtracting equation (1) from (2),
26,810 - 2904 = 2a + 19d - (2a + 9d)
23,906 = 10d
d = 2390.6
Put this value in equation (1).
2904 = 2a + 9(2390.6)
2904 = 2a + 21515.4
2904 - 21515.4 = 2a
a = -9305.7

Here, a = first term, d = common difference
51^st term = 5(15^th term)
a + (51 - 1)d = 5(a + (15 - 1)d)
a + 50d = 5(a + 14d)
a + 50d = 5a + 70d
50d - 70d = 5a - a
-20d = 4a
a + 5d = 0

Here, a = first term, d = common difference, S_n = sum of n number of terms, n = number of terms
16^th term = -19
a + (16 – 1)d = -19
a + 15d = -19 --- (1)
Sum of the first 11 terms = 77
n/2 (2a + (n – 1)d) = 77
(11/2)(2a + 10d) = 77
2a + 10d = 14 --- (2)
Multiplying equation (1) by 2;
2(a + 15d = -19)
2a + 30d = -38 --- (3)
Subtracting equation (3) from (2);
2a + 10d – 2a – 30d = 14 - (-38)
-20d = 52
d = -2.6
Use the value of d in equation (1).
a + 15(-2.6) = -19
a – 39 = -19
a = -19 + 39 = 20
Sum of the first 20 terms =n/2 (2a + (n – 1)d)
= 20/2 (2 × 20 + 19 × - 2.6)
= 10(40 – 49.4) = - 94