I think you talk about this question Insert the missing number. 2, 6, 12, 20, 30, 42, 56, (….) A) 61 B) 64 C) 72 D) 70 Answer: C) 72 Explanation: The pattern is 1 x 2, 2 x 3, 3 x 4, 4 x 5, 5 x 6, 6 x 7, 7 x 8. So, the next number is 8 x 9 = 72. Reply
1^1+1=2 2^2+2=6 3^2+3=12 4^2+4=20 5^2+5=30 nth term=n^2+n Next number=6^2 +6=36+6=42 II method: 1*(1+1)=2 2*3=6 3*4=12 4*5=20 5*6=30 nth term=n*(n+1) next no=6th term= 6*(6+1)=6*7=42 EDIT 1: III METHOD: 0+2*1=2 2+2*2=6 6+2*3=12 12+2*4=20 20+2*5=30 Thus nth term Tn=Tn-1+2*n Or Tn-Tn-1=2*n So we can write T2-T1=2*2 T3-T2=2*3 T4-T3=2*4 ……………….. ……………….. Tn-1-Tn-2=2*(n-1) Tn-Tn-1=2*n ———————————————— Adding all we get Tn-T1=2*(2+3+4+………………………….+n) Tn=2*(1+2+3+4+………………………….+n) =2*n(n+1)/2=n(n+1) Tn=n(n+1) which is same as above Thus T6=T5+2*6=30+12=42 METHOD IV: Tn=(n^3–1) /(n-1) – 1 Thus T3=(27–1)/(3–1)-1=26/2–1=13–1=12 T4=(64–1)/(4–1) -1 =63/3–1=21–1=20 T5=(125–1) / ( 5–1)-1=124/4–1=31–1=30 Next term=T6=(6^3–1)/(6–1) -1=(216–6)/(6–1) -1=215/5–1=43–1=42 EDIT 2: METHOD V: nth term=(n+1)(n-1) +n+1 T1=0*2+2=2 T2=1*3+3=6 T3=2*4+4=12 T4=3*5+5=20 T5=4*6+6=24+6=30 Next term T6=5*7+7=35+7=42 EDIT 3: METHOD VI: nth term=(n-1)^2 +3n-1 T4=(4–1)^2+3*4–1=9+12–1=20 T5=(5–1)^2 +3*5–1=16+15–1=30 Next term T6=(6–1)^2 +3*6–1=25+18–1=42 Reply
What is the missing number in the series 2, 6, 12, 20, 30, ..? A. 72 B. 42 C. 52 D. 62 E. 56 so 2+4=6 6+6=12 12+8=20 20+10=30 30+12=42 Reply
42
72
I think you talk about this question
Insert the missing number.
2, 6, 12, 20, 30, 42, 56, (….)
A) 61 B) 64
C) 72 D) 70
Answer: C) 72
Explanation:
The pattern is 1 x 2, 2 x 3, 3 x 4, 4 x 5, 5 x 6, 6 x 7, 7 x 8.
So, the next number is 8 x 9 = 72.
1^1+1=2
2^2+2=6
3^2+3=12
4^2+4=20
5^2+5=30
nth term=n^2+n
Next number=6^2 +6=36+6=42
II method:
1*(1+1)=2
2*3=6
3*4=12
4*5=20
5*6=30
nth term=n*(n+1)
next no=6th term= 6*(6+1)=6*7=42
EDIT 1:
III METHOD:
0+2*1=2
2+2*2=6
6+2*3=12
12+2*4=20
20+2*5=30
Thus nth term
Tn=Tn-1+2*n Or Tn-Tn-1=2*n
So we can write
T2-T1=2*2
T3-T2=2*3
T4-T3=2*4
………………..
………………..
Tn-1-Tn-2=2*(n-1)
Tn-Tn-1=2*n
————————————————
Adding all we get
Tn-T1=2*(2+3+4+………………………….+n)
Tn=2*(1+2+3+4+………………………….+n)
=2*n(n+1)/2=n(n+1)
Tn=n(n+1)
which is same as above
Thus T6=T5+2*6=30+12=42
METHOD IV:
Tn=(n^3–1) /(n-1) – 1
Thus T3=(27–1)/(3–1)-1=26/2–1=13–1=12
T4=(64–1)/(4–1) -1 =63/3–1=21–1=20
T5=(125–1) / ( 5–1)-1=124/4–1=31–1=30
Next term=T6=(6^3–1)/(6–1) -1=(216–6)/(6–1) -1=215/5–1=43–1=42
EDIT 2:
METHOD V:
nth term=(n+1)(n-1) +n+1
T1=0*2+2=2
T2=1*3+3=6
T3=2*4+4=12
T4=3*5+5=20
T5=4*6+6=24+6=30
Next term T6=5*7+7=35+7=42
EDIT 3:
METHOD VI:
nth term=(n-1)^2 +3n-1
T4=(4–1)^2+3*4–1=9+12–1=20
T5=(5–1)^2 +3*5–1=16+15–1=30
Next term T6=(6–1)^2 +3*6–1=25+18–1=42
42
What is the missing number in the series 2, 6, 12, 20, 30, ..?
A. 72 B. 42 C. 52 D. 62 E. 56
so 2+4=6
6+6=12
12+8=20
20+10=30
30+12=42
The answer is 42