Sequences and Series – MBA CET questions with solutions

Approach: AP sum of multiples in a range.

Solution:
First multiple of 6 above 100 = 102; last below 400 = 396.
Number of terms = (396 − 102)/6 + 1 = 50.
Sum = 50/2 × (102 + 396) = 25 × 498 = 12,450.

Cetking shortcut: Sum = 6 × (17 + 18 + … + 66) = 6 × 50 × 83/2 = 12,450.

Answer: C

Approach: Three numbers in GP: take them as a/r, a, ar so the product is a³.

Solution:
Product a³ = 216 → a = 6.
Sum: 6/r + 6 + 6r = 26 → 6/r + 6r = 20 → 3r² − 10r + 3 = 0.
r = 3 or 1/3 → numbers 2, 6, 18.
Check: 2 + 6 + 18 = 26, 2 × 6 × 18 = 216. ✓

Why the other options are wrong:

• A, B: 4, 6, 16 sums to 26 but 6/4 ≠ 16/6 (not a GP); product 384.
• C: 9, 6, 11 is not a GP; product 594.
• E: 5, 7, 14 is not a GP; product 490.

Cetking shortcut: Middle term of a 3-term GP = cube root of the product = 6; then check which option has 6² = first × third (2 × 18 = 36).

Answer: D

Approach: Sum of an AP = number of terms × average of first and last term.

Solution:
Number of terms = 86 − 37 + 1 = 50
Average = (37 + 86)/2 = 123/2 = 61.5
Sum = 50 × 61.5 = 3075

Cetking shortcut: 50 × 61.5 = 25 × 123 = 3075 — pairing terms (37+86 = 123, 25 pairs) gives the same result.

Answer: A

Approach: Sum of first n natural numbers = n(n + 1)/2.

Solution:
n = 60
Sum = 60 × 61 / 2 = 30 × 61 = 1830

Cetking shortcut: Sum must be a multiple of 61 (and of 30): only 1830 = 30 × 61.

Answer: E

Approach: Sum of first n even natural numbers = n(n + 1).

Solution:
2 + 4 + … + 200 = 2(1 + 2 + … + 100) = 2 × 100 × 101/2 = 100 × 101 = 10100.

Why the other options are wrong:
D 10050: uses n(n+1)/2 × 2 wrongly or 100 × 100.5.
B, C, E: arithmetic slips.

Cetking shortcut: n(n + 1) = 100 × 101 = 10100.

Answer: A

Approach: Three numbers in AP: take them as a − d, a, a + d.

Solution:
Sum = 3a = 48 ⇒ a = 16.
Sum of squares = 3a² + 2d² = 800 ⇒ 768 + 2d² = 800 ⇒ d² = 16 ⇒ d = 4.
Numbers: 12, 16, 20 ⇒ smallest = 12.

Why the other options are wrong:
A 10 ⇒ 10, 16, 22: squares sum 840.
C 11, D 9, E 13: squares sum 818, 866, 786 — not 800.

Cetking shortcut: Middle term = 48/3 = 16; test options: 12, 16, 20 ⇒ 144 + 256 + 400 = 800 ✓.

Answer: B

Approach: Sum of first n even natural numbers = n(n + 1).

Solution:
Even numbers: 2 + 4 + … + 180 = 2(1 + 2 + … + 90)
= 2 × 90 × 91/2 = 90 × 91
= 8190

Cetking shortcut: n(n+1) directly: 90 × 91 = 8190. Option D (8100 = 90²) is the sum of the first 90 odd numbers — a trap.

Answer: B

Approach: Pair up terms that add to 470; take at most one from each pair.

Solution:
Number of terms = (467 − 3)/8 + 1 = 59.
Pairs adding to 470: (3, 467), (11, 459), … — the middle term is 235 (3 + 8 × 29), and 235 + 235 = 470 needs two equal elements, so 235 has no partner.
Pairs = (59 − 1)/2 = 29. Take one from each pair + 235 → 29 + 1 = 30.

Cetking shortcut: Odd number of terms symmetric about 235: max = (n + 1)/2 = 30.

Answer: E

Approach: Total balls in n triangular layers = tetrahedral number n(n + 1)(n + 2)/6.

Solution:
n(n + 1)(n + 2)/6 = 8436 → n(n + 1)(n + 2) = 50616.
∛50616 ≈ 37 → try 36 × 37 × 38 = 50616 ✓
So n = 36 layers. (Check n = 38: 38 × 39 × 40/6 = 9880 ≠ 8436.)

Cetking shortcut: 8436 × 6 = 50616 ends in 6; 36 × 37 × 38 ends in 6 (6 × 7 × 8 = 336), 38 × 39 × 40 ends in 0 — so 36.

Answer: C

Approach: Place a, b, c, d on a number line, then fit e.

Solution:
Take a = 0, b = 4k (k ≠ 0). c = mean of a, b = 2k; d = mean of b, c = 3k
0, 2k, 3k, 4k plus e must be 5 equally spaced values → e = k (common gap k)
So a = 0, e = k, c = 2k, d = 3k, b = 4k
i. Average = (0 + k + 2k + 3k + 4k)/5 = 2k = c → true
ii. Avg(d, e) = 2k, Avg(a, b) = 2k → 'not greater' → true
iii. Avg(b, c) = 3k vs Avg(a, d) = 1.5k → greater only if k > 0; for a decreasing AP (k < 0) it fails → not necessarily true

Why the other options are wrong:
B, C, E: treat iii as always true, but it reverses for a decreasing AP.
D: misses statement i, which holds for every k.

Cetking shortcut: Equalities and 'not greater' (here an equality) survive a sign flip of k; a strict 'greater' comparison does not.

Answer: A

Approach: Sum of triangular numbers = n(n+1)(n+2)/6.

Solution:
Layers: 1, 3, 6, 10, … = n(n+1)/2
Total = n(n+1)(n+2)/6 = 8436
n(n+1)(n+2) = 50616
Cube root of 50616 ≈ 37 → try 36 × 37 × 38 = 50616 ✓
n = 36 layers

Why the other options are wrong:
A (32), D (34): give 5984 and 7140 balls.
E (24): 2600 balls.
C (96): far too many.

Cetking shortcut: n³/6 ≈ 8436 → n³ ≈ 50600 → n ≈ 37 → the middle of three consecutive numbers is 37, so n = 36.

Answer: B