Arithmetic Progressions Test - 2

a₁ = 56 = 7 × 8
a_n = 497 = 71 × 7
So, S_n = 56 + 63 + 70 + ............ + 497
= 7(8 + 9 + 10 + ........... + 71)
Numbers of terms = 71 - 8 + 1 = 64
S_n = (64/2) × 7 × (8 + 71)
= 7 × 79 × 32 = 17696

Let the AP be a, a + d, a + 2d, …
∵ T_n = a + (n - 1)d
T₃ + T₁₅ = T₆ + T₁₁ + T₁₃
2a + 2d + 14d = 3a + 5d + 10d + 12d
0 = a + 11d
0 = T₁₂
 

a + (m - 1) d = n ............(1)
a + (n - 1) d = m ............(2)
By subtracting (2) from (1), we get (m - n) d = (n - m)
d = - 1 .............(3)
By substituting the value of d from equation (3) into equation (1), we get a + (m - 1)( - 1) = n
a = n + m - 1
T_{m + n} = a + (m + n - 1) d = m + n - 1 + (m + n - 1) (- 1) = 0

Here, a = first term, d = common difference, S_n = sum of n number of terms, t_n = n^th term
The numbers of plants in the field rows form an A.P.
a = 80
d = 76 - 80 = -4
Let the number of rows be n.
t_n = a + (n - 1)d
So, 24 = 80 + (n – 1)(-4);
24 = 80 - 4n + 4
24 = 84 - 4n
24 - 84 = - 4n
-60 = -4n
Or n = 15

Here, a = first term, d = common difference, S_n = sum of n terms, n = number of terms, t_n = n^th term
d = -170 - (-200) = -170 + 200 = 30, n = 50
Sum of n terms = n/2(2a + (n – 1)d)
= 50/2([2 × -200] + (50 - 1) × 30)
= 25(-400 + 49 × 30)

= 25(-400 + 1470)

= 25(1070)
S₅₀ = 26,750

n = 20
Sum of n terms = n/2(2a + (n – 1)d)
= (20/2)([2 × -200] + (20 -1) × 30)
= 10(-400 + 19 × 30)
= 10(-400 + 570)
= 10(170)
S₂₀ = 1700

Here, a = first term, d = common difference, S_n = sum of n number of terms, t_n = n^th term
A.P.₁ = 28, 26.5, 25
First term of the first A.P. = a₁ = 28
Common difference of the first A.P. = d₁ = 26.5 - 28 = -1.5
A.P.₂ = 4, 4.5, 5, ...
First term of the second A.P. = a₂ = 5
Common difference of the second A.P. = d₂ = 4.5 - 4 = 0.5
According to the question, n^th terms of these A.P.s are equal.
a₁ + (n - 1)d₁ = a₂ + (n - 1)d₂
28 + (n - 1)(-1.5) = 4 + (n - 1)0.5
28 - 1.5n + 1.5 = 4 + 0.5n - 0.5
29.5 - 1.5n = 3.5 + 0.5n
29.5 - 3.5 = 0.5n + 1.5n
26 = 2n
n = 26 ÷ 2
n = 13