Which letter cluster will replace the question mark (?) to complete the given series? abCD, fgHI, ?, stUV, abCD
lmNO
The question asks us to find the letter cluster that completes the given series: abCD, fgHI, ?, stUV, abCD.
Let's analyze the pattern by looking at how the letters change from one term to the next. We can observe the change in the alphabetical position of each letter. It's also important to notice that the first two letters are lowercase and the last two are uppercase in every term.
Let's look at the alphabetical position of the letters in the first two terms:
abCD: a(1), b(2), C(3), D(4)fgHI: f(6), g(7), H(8), I(9)Now, let's calculate the difference in position for each corresponding letter:
So, from the first term to the second term, each letter's position increased by 5. Let's call this increment $\Delta_1 = +5$.
Now let's look at the gap between the last two terms provided, which are stUV and abCD. Remember that the alphabet wraps around after Z (position 26).
stUV: s(19), t(20), U(21), V(22)abCD: a(1 or 27), b(2 or 28), C(3 or 29), D(4 or 30)Let's calculate the difference in position, considering the wrap-around:
The increment from the fourth term to the fifth term (which is the same as the first term) is $\Delta_4 = +8$.
We have increments of +5 (from term 1 to 2) and +8 (from term 4 to 5). This suggests that the increments between terms might be increasing in a sequence. Let's hypothesize the sequence of increments is +5, +6, +7, +8. This means:
Let's test this hypothesis to find the missing term (Term 3).
According to our hypothesis, the missing term (Term 3) is obtained by adding an increment of +6 to each letter of Term 2 (fgHI).
fgHILet's calculate the letters for the third term:
Combining these letters, we get lmNO.
Now let's check if applying the next increment ($\Delta_3 = +7$) to our calculated term lmNO gives us the fourth term stUV.
lmNOLet's calculate:
The pattern holds true. The sequence of increments between consecutive terms is +5, +6, +7, +8, and then it cycles back to +5 (as seen from the last term returning to the first).
The missing term in the series abCD, fgHI, ?, stUV, abCD is obtained by applying an increment of +6 to each letter of the previous term fgHI. This results in lmNO.
The completed series is: abCD, fgHI, lmNO, stUV, abCD.
Let's look at the options provided:
Our calculated term lmNO matches Option 2.
| From Term | To Term | Increment per Letter |
|---|---|---|
| abCD | fgHI | +5 |
| fgHI | lmNO | +6 |
| lmNO | stUV | +7 |
| stUV | abCD | +8 (with wrap-around) |
| Concept | Explanation | Example Application |
|---|---|---|
| Alphabetical Position | Each letter corresponds to a number (A=1, B=2, ..., Z=26). This helps in quantifying the differences between letters. | Finding the difference between 'c' and 'g': g(7) - c(3) = +4. |
| Constant Increment | The same number is added or subtracted to get the next letter in the series (e.g., A, C, E, G... where +2 is added each time). | B, E, H, K (B+3=E, E+3=H, H+3=K) |
| Variable Increment | The number added or subtracted changes according to a specific pattern (e.g., +1, +2, +3... or +5, +6, +7...). | A, C, F, K (A+2=C, C+3=F, F+5=K - gaps are prime numbers) |
| Wrap-around | When increments go past Z (26), the count continues from A (27=A, 28=B, etc.). Similarly, counting backward from A wraps around to Z. | V(22) + 8 = 30. 30 mod 26 = 4, which is D. |
| Multiple Series | Sometimes, a single series is formed by interleaving two or more simpler series. | A, P, B, Q, C, R (alternating A,B,C and P,Q,R) |
Letter series questions are common in logical reasoning tests. To solve them effectively, consider these tips:
Practice is key to becoming proficient in identifying different types of letter series patterns.
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