Test Level 1: Progressions, Sequences & Series - 1 CAT MCQs solutions -

The n^th term of the given series is t_n = n² + n, n = 1, 2, 3, ..., n

If $a is the first installment and the common difference is $d,
Then,
Sum of n terms

Sum of all installments 

= total debt = 3,600
⇒ 2a + 39d = 180 …(1)

of the total debt = 2400.
⇒ 2a + 29d = 160 …(2)
Subtracting (2) from (1), we get, 10d = 20.
∴ d = 2
∴ a = $51

50th term =
Then, 

The given series can be written as:

If a, b, c and d are in GP, then
b² = ac, c² = bd and ad = bc.
Given equation is:
a² + c² - 2ac + b² + c² - 2bc + b² + d² - 2bd
= [a² + 2b² + 2c² + d² - 2ac - 2bc - 2bd]
= [a² + d² + 2ac + 2bd - 2ac - 2bc - 2bd]
= [a² + d² - 2ad] = (a - d)²

t₉ = t₆ + 5
⇒ a + 8d = a + 5d + 5
⇒ d = 5/3
t₁₁ = a + 10d = 35

x, y, z are in AP.
∴ 2y = (z + x)
Add (z + x) both sides.
2y + (z + x) = 2(z + x)
⇒ 2(z + x) = (y + z) + (x + y)
⇒ (y + z), (z + x), (x + y) are in AP.

Let the numbers be a - 2d, a - d, a, a + d, a + 2d.
Then, 5a = 75
a = 15
(a - 2d).(a + 2d) = 161
This implies, d = +4
Greatest term = 15 + 2 x 4 = 23
The greatest term would be the 5th term, if d = +4, and the 1st term, if d = -4.

T₉ + T₁₄ + T₁₈ = T₂₀ + T₂₄
⇒ 3a + 38d = 2a + 42d
⇒ a = 4d