Test Level 2: Progressions, Sequences & Series - 1 - CAT MCQ

Given: a = 42°, d = 33°
Sum of all the interior angles of an n-sided polygon is (n - 2)180°.

(n - 2) × 180 = (n/2)(2 × 42 + (n - 1) × 33)
360n - 720 = 84n + 33n² - 33n
11n² - 103n + 240 = 0
(n - 5)(11n - 48) = 0
This implies, n = 5 as sides can't be fraction in number.

Since the sum to infinity of the given GP exists, so its common ratio should be 0 < |3/x| < 1.
0 < 3/x is always true because x > 0.
3/x < 1
or

or
x > 3
According to question, x is not negative.
So, x > 3

x, y, z are in A.P.
Therefore, 2y = x + z
Given equation: (x + 2y - z)(2y + z - x)(z + x - y)
= (x + x + z - z)(x + z + z - x)(2y - y)
= 2x(2z)(y) = 4xyz

T₃ - T₂ = ar² - ar = 10
⇒ ar(r - 1) = 10
and T₂ - T₁ = 6, or ar - a = 6
⇒ a(r - 1) = 6
r = 10/6 = 5/3 and a = 9

4a + 14d = 2a + 23d

Total number of chocolates 

Let the GP be a, ar, ar², ar³, ………….

Since log₃ (2), log₃ (2x - 5) and log₃  are in AP, therefore we can write:

∴ log3 (2x - 5)2 = log3 (2x + 1 - 7)
∴ (2x - 5)2 = 2x + 1 - 7
Let, 2x = y.
Then, (y - 5)2 = 2y - 7
⇒ y2 + 25 - 10y = 2y - 7
⇒ y2 - 12y + 32 = 0
⇒ (y - 8)(y - 4) = 0
We get, 2x = 8 and 2x = 4
⇒ x = 3 and 2
If x = 2, then 2x - 5 = 4 - 5 = -1< 0, not possible
So x = 3

= 50/51

The nth term of the given series is t_n 

The sum of the series is