Test: Sequence & Series - 2 - CAT MCQ

The given series is a GP with common ratio 1/4. Hence, option (a) is correct.

Obviously 30 is the misplaced number because all other terms are in a reducing series here. Option (b) is correct.

The logic followed in this series is 4 × 5 + 6 = 26; 26 × 6 +7 = 163; 163 × 7 + 8 =1149; 1149 × 8 + 9 = 9201.
Only the last number in the series (82809) breaks this trend.
Instead the correct value should have been 9201 × 9 + 10 = 82819.
Thus, Option (c) is correct.

The pattern is:
2³ + 1² = 9
3³ + 2² = 31
4³ + 3² = 73
5³ + 4² = 141
6³ + 5² = 241

The pattern is:
658 + 72 = 730
730 + 144 = 874
874 + 288 = 1162
1162 + 576 = 1738

S_n - S_n = t_n
Substitute m instead of n
S_m - S_m-1 = t_m
We know that S_n = n³ + n² + n + 1
Hence m³ + m² + m + 1
m³ + m² + m + 1 - [(m-1)³ + (m-1)² + (m-1) + 1 ] = 291
m³ + m² + m + 1 - [m³ - (3m)² + 3m - 1 + m² - 2m + 1 + m - 1 + 1] = 291
1 + (3m)² + 3m + 1 -2m - 1 = 291
(-3m)² + m - 290 = 0
(3m)² - m + 290 = 0
Solving above equation we get m = -29, 30
M cannot be negative
Hence m = 30

Given, Second term of an AP is 8 => a+d =8, 8th term is 2 more than thrice the second term
=> a+7d = 2 +3(a+d) = 2+3*8 =26 .
a+d = 8 -------------- 1
a+7d = 26 -------------- 2
Solving for d in the two equations, we get d = 3 and a =5.
Sum of n terms in an AP = n/2 * [2a +(n-1)d].
=> Sum upto 8 terms in this AP = 8/2 * [2*5 + (8-1)3] => 4*[10+21] = 4*31 = 124.

A: 15, 19, 23, 27, . . . . , 415

B: 14, 19, 24, 29, . . . , 464

Here the first common term = 19

Common difference = LCM of 5, 4 = 20
19+(n-1)20 ≤ 415
(n-1)20 ≤ 396
(n-1) ≤ 19.8
n=20

The third term t₃ = a + 2d
The ninth term t₉ = a + 8d
t₃ + t₉ = 2a + 10d = 8
Sum of first 11 terms of an AP is given by
⇒ S₁₁ = 11/2 [2a + 10d]
⇒ S₁₁ = 11/2 × 8
⇒ S₁₁ = 44

46080 ÷ 12 = 3840
3840 ÷ 10 = 384
384 ÷ 8 = 48
48 ÷ 6 = 8
8 ÷ 4 = 2
2 ÷ 2 = 1