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

S = 3x² + 5x³ + 7x⁴ + .............(1)
Sx = 3x³ + 5x⁴ + 7x⁵ + ...............(2)
Subtracting (2) from (1), we get
S(1 - x) = 3x² + 2x³ + 2x⁴ + 2x⁵ + .........

Sum of 20 terms 


= 10 (2.6 + 41.8)
= 10 × 44.4 = 444

In the given series, odd terms are in AP and even terms are in GP.
(1 + 5 + 9 + …….. 5 terms) + (3 + 6 + 12 + ………. 5 terms)
= 45 + 93 = 138.

Given: a, b, c are in AP. Divide all the terms by abc, resulting sequence will also be in AP.

3 + 6 + 10 + 16 + …… n terms
= (1 + 2) + (2 + 4) + (4 + 6) + (8 + 8) + … n terms
= (1 + 2 + 4 + 8 + ... n terms) + (2 + 4 + 6 + 8 + … n terms)

a₂ – a₁ = a₃ – a₂ = ........... = d ≠ 3m
a₁, a₁ + d, a₁ + 2d, a₁ + 3d, ….
d = 3λ + 1, 3λ + 2
a₁, a₁ + 1 + 3λ, a₁ + 2 + 6λ, a₁ + 3 + 9λ, ….
a₁, a₁ + 2 + 3λ (prime), a₁ + 4 + 6λ, a₁ + 6 + 9λ, …. 0
Hence, Infinite

Here, t₃ = 600 and t₇ = 700.
i.e. a + 2d = 600 ...........(1)
and a + 6d = 700 ...........(2)
Solving (1) and (2),
d = 25 and a = 550
(i) a, the production in the first year = 550
(ii) t10, the production in the 10th year = a + 9d = 550 + 9 (25) = 550 + 225 = 775
(iii) total production in 7 years

2 - 2r = 1 + r

a = 2
Therefore, a³ + a³r³ + a³r⁶ + ....

So, option (2) is the answer.

Given: a_m = 1/m
⇒ a + (m - 1)d = 1/n
⇒ an + mnd - nd = 1 ... (1)
a_m = 1/m
⇒ a + (n - 1)d = 1/m
⇒ am + mnd - md = 1 ... (2)
From (1) and (2), we get
an + mnd - nd = am + mnd - md
⇒ a(n - m) - (n - m)d = 0
⇒ a = d
Consider (1): an + mnd - nd = 1
dn + mnd - nd = 1
d = 1/mn
Hence, a = 1/mn

Here, first term, a = 2 … (1) 
Fourth term = a + 3d = 6 ... (2)
∴ 2 + 3d = 6

⇒ 6800 × 2 × 3 = 8n + 4n²
⇒ 40,800 = 8n + 4n²
⇒ 4n² + 8n - 40,800 = 0
⇒ n² + 2n - 10,200 = 0
⇒ (n + 102)(n - 100) = 0
⇒ n = 100 or n = -102 (not possible)
∴ n = 100

Suppose that a and d be the first term and common difference of an AP.
S₈ = 8/2 [2a + (8 - 1)d] = 64
⇒ 2a + 7d = 16 …… (1)
Also, S₁₉ = 19/2 [2a + (19 - 1)d] = 361
⇒ 2a + 18d = 38 ..…. (2)
Subtracting (1) from (2), we get:
11d = 22
∴ d = 2
∴ 2a = 16 - 14 = 2
∴ a = 1
S_n = n/2 [2a + (n - 1)d] =n/2 [2 + (n - 1)2]
= 2n + n² - 2n = n²

Let the harmonic progression be: 1/a, 1/(a + d), 1/(a + 2d), …
Therefore, we have a + (a + d) + (a + 2d) + … + (a + 8d)
= (9/2)[(a) + (a + 8d)]
= 9(a + 4d)

S₁₁ = S₁₉
This means that the sum of terms from T₁₂ to T₁₉ is zero; which indicates that out of these 8 terms, half are positive and half are negative with same magnitude. Also, the next 11 terms have the same sum as the first 11 terms are having, but of opposite sign. This will result in the sum of the first 30 terms as 0 (zero).

The series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, ...
1, 2, 3, 4, 5 and so on.
Sum of n integers starting from 1 is given by:
Sum of n integers starting from 1 is given by:

n₁(n₁ + 1) < 576
If n₁ = 24, LHS < 576
Thus, for n₁ = 23.

Thus, n₁ = 24 will start the series from 277th term.
Also, n₁ = 24 corresponds to 'x'.

The required terms are: 33, 44, …, 528. They form an AP with a common difference of 11.
528 = 33 + (n - 1)11
⇒ n - 1 = 495/11 = 45
⇒ n = 46

Clearly, the numbers between 250 and 1000 which are divisible by 3 are: 252, 255, 258, …, 999.
This is an AP with first term, a = 252, common difference = 3, and last term = 999.
Let there be n terms in this AP.
Then an = 999
⇒ a + (n - 1)d = 999
⇒ 252 + (n - 1) × 3 = 999
⇒ n = 250
∴ Required sum = S_n =  n/2 [a+1]
= 250/2 [252 + 999] = 1,56,375

Let the first term of the AP be a and the common difference be d.
S₄ = 28 = (4/2)(2a + 3d) = 2(2a + 3d)
Or 2a + 3d = 14 ... (i)
S₈ = 48 = (8/2)(2a + 7d) = 4(2a + 7d)
Or 2a + 7d = 12 ... (ii)
Now, subtracting (i) from (ii), we have
4d = -2
d = (-1/2) ...(iii)
Putting in (i), we get
a = (31/4)
S₁₂ = (12/2)(2a + 11d)
= 6(2a + 11d)
= 12a + 66d

= 93 - 33
= 60
Hence, answer option 3 is correct.

Let the numbers be:
a - 2d, a - d, a, a + d, a + 2d
Given that the sum = 30
a - 2d + a - d + a + a + d + a + 2d = 30
5a = 30
a = 6
(6 - 2d)² + (6 - d)² + (6)² + (6 + d)² + (6 + 2d)² = 220
10d² + 180 = 220
d² = 4d = ±2
Hence, the numbers are 2, 4, 6, 8 and 10.

Numbers of balls in horizontal layers:
1, 3, 6, 10, …
in which each nth term is the sum of the first `n` natural numbers.

If there are m layers, then we should have


If we substitute the options one by one in the above equation, we see that m = 36 satisfies it.