Series Test - 1

a + 5d = 19 ……(i)
a + 16d = 41 ….… (ii)
Subtracting (i) from (ii), we get
11d = 22 ⇒ d = 2
From (1), a + 5 × 2 = 19 ⇒ a = 9
Now 40th term = a + 39d = 9 + 39 × 2 = 9 + 78 = 87

As the difference between 2 consecutive terms is the same, this is an AP.
1^st term of this AP = a = 44
Common difference = d = - 4
Now, the sum will be maximum, if all the terms are positive.
Let the n^th term be the smallest positive term of this AP.
Thus, T_n = a + (n - 1)d = 44 + (n - 1)(- 4) ≥ 0
Or 44 + (n - 1)(- 4) ≥ 0
Or 48 - 4n ≥ 0
Thus, 12 ≥ n
Thus, for maximum sum, n = 12
S_n = (n/2)(2a + (n - 1)d)
Putting the values n = 12, a = 44, and d = - 4 in the above equation, we get
S₁₂ = 264
Thus, answer option 2 is correct.

Let 4 terms be a, b, c, d.
Sum of extremes = Sum of means
So, a + d = b + c = 16
And ad = 28
So, a = 14, d = 2
Or d = 14, a = 2

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
Tm + n = a + (m + n - 1) d = m + n - 1 + (m + n - 1) (- 1) = 0

Tn = 2n - 5, Putting n = 1, 2, 3 ……, the series is - 3, - 1, 1, 3, 5, 7, 9, 11, 13, 15
Thus, the sum of the first five terms to the sum of the next five terms = 5/55 or 1 : 11.

Common difference = (4K - 6) - (K + 2) = (3K - 2) - (4K - 6)
3K - 8 = - K + 4
4K = 12
K = 3