Question

find the next element in the following sequence P7B16,T14C20,X21D24,?​

Answer 1

Answer:

B28E2

Step-by-step explanation:

first digit is the alphabet number of given last digits and last two digits difference is 4

Answer 2

Answer:

We can find the last element in the series using the simple logic.

The Final Answer is: $P7B16,T14C20,X21D24,B28E28$

Step-by-step explanation:

First we write the Alphabets:

A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z

Now we compare the elements in the given series. Take away the terms in the series.

First term $=P7B16$ , second term $=T14C20$ , third term $=X21D24$

• Three alphabets are between $P$ and $T$ ⇒ (Q, R, S).

• Three alphabets are between $T$ and $X$ ⇒ (U, V, W).

Now we can find the first value of last term.

• Three alphabets are between $X$ and $B$. First value of last term $=B$

Next we find the second value of last term.

• First term has the value $7=(1*7)$
• Second term has the value $14=(2*7)$
• Third term has the value $21=(3*7)$.

These Values are multiple of $7$. Second value of last term is $28=(4*7)$

Next we find the third value of last term.

• First term has the alphabet $B$
• Second term has the alphabet $C$
• Third term has the alphabet $D$

These are the alphabets sequence. So that Third value of last term is $E$

Next we find the fourth value of last term.

• First term has the value $16$
• Second term has the value $20=(16+4)$
• Third term has the value $24=(20+4)$

Similarly we should add the value $4$ and $24$ ⇒ $(28=24+4)$.  Fourth value of last term is $28$.

Final Answer: $B28E28$