What is the missing number in the series 2, 6, 12, 20, 30, ..?

7 Answers on this Question

1. 42

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2. 72

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   1. 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.

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3. 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

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4. 42

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5. 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

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6. The answer is 42

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