What is the next number of the following sequence 3, 14, 39, 84, ?

Kulbeer Singh Sidhu   10 years AGO

3=1*3
14=2*7
39=3*13
84=4*21
?=5*(n)
7-3=4, 13-7=6, 21-13=8, n-21=10 =>n=31
5*31=155

Like? Yes (21) | No (10)  

prem kumar   10 years AGO

1*3=3
2*7=14
3*13=39
4*21=84
5*31=155

Like? Yes (9) | No   

nisha sharma   10 years AGO

as difference between multiple is in series 4,6,8
3*1=3
7*2=14
13*3=39
21*4=84
31*5=155

Like? Yes (10) | No (2)  

Niveditha Ravichandran   10 years AGO

n^3 + n^2 + n

n=1, 1+1+1=3
n=2, 8+4+2 =14
n=3, 27+9+3 =39
n=4, 64+16+4 =84
n=5, 125+25+5 =155

Like? Yes (6) | No   

sanjeev kumar   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84= 4 * 21

3,7,13,21...
7-3=4
13-7=6
21-13=8
_ -21=10
so next number 21+10=31
and 5*31=155

Like? Yes (6) | No   

suraj tiwary   10 years AGO

the general term of this series is n^3+n^2+n..so for n=5...the term will be..125+25+5=155

Like? Yes (4) | No   

Amit Kumar Pandit   10 years AGO

3*1=3
7*2=14
13*3=39
21*4=84
31*5=155

Like? Yes (4) | No   

posirajesh   10 years AGO

use nth number =n×[n*n+(n+1)]

Like? Yes (4) | No   

Venkateswarlu   10 years AGO

14-3=11
39-14=25 25-11=14
84=39=45 45-25=20 14,20,26,-,-,--
39=11+14+14=39
84=25+39+20
// 45+84+26=155

Like? Yes (3) | No   

ESHA CHATTERJEE   10 years AGO

1st no.=(1^2+2)*1=3
2nd no.=(2^2+3)*2=14
3rd no.=(3^2+4)*3=39
4th no=(4^2+5)*4=84
5th no=(5^2+6)*5=155

Like? Yes (3) | No   

Neel Patel   10 years AGO

3=1+1^2+1^3
14=2+2^2+2^3
36=3+3^2+3^3
84=4+4^2+4^3
so
5+5^2+5^3=155

Like? Yes (3) | No   

ashok   10 years AGO

s= 3, 14, 39, 84, ?
-s= 3, 14, 39, 84, A
-------------------------------------
s1=3, 11 25 45 B
-s1= 3, 11 25 45
---------------------------------------
s2=3, 8, 14, 20 C
-s2= 3, 8, 14 20
------------------------------------
s4=3, 6, 6, 6 D

So D=6
C=D+20=26
B=C+45=71
A=B+84=155

Ans=155

Like? Yes (4) | No (1)  

lebaka pavan   10 years AGO

Given numbers 3,14,39,84,?
Difference between given numbers are 11,25,45,?
Then the difference between above numbers are 14,20,?
Therefore,the difference is 6
So the next number for the above difference is 26
So the next number is 26+45+84=155

Like? Yes (3) | No   

Tanya Kumar   10 years AGO

basic series

Like? Yes (3) | No   

ravikumar   10 years AGO

let us take the series in quadratic form
an^2 + bn +c
replace a=1 n=1 1(1)+1(1)+1=3
a=2 n=2 2(4)+2(2)+2=14
similarly
a=4 n=4 4(16)+4(4)+4=84
when a=5 n=5 5(25)+5(5)+5= 155

hence 155 is correct

Like? Yes (3) | No   

Sourojit Dhua   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

answer = option 9, 155

Like? Yes (8) | No (6)  

sumit   10 years AGO

series is n^3+n*(n+1)
1^3+1*(1+1)=3
2^3+2*(2+1)=14
so the 5th term will be
5^3+5*(5+1)=155 opt(9) ans

Like? Yes (7) | No (5)  

SAYAN CHATTERJEE   10 years AGO

3+11=14
14+25=39
39+45=84
differences are 11, (11+14=25), (25+20=45)
so, next difference is (45+26=71).
so, next no. is (45+71=155) (9)

Like? Yes (2) | No   

urvish   10 years AGO

here 3 = 1 * 3,14 = 2 * 7,39 = 3 * 13,84 = 4 * 21.....
T(n) = n[(n+1)^2 - n]
so next n=5
=5*((5+1)^2-5)
=5*(36-5)
=5*31
=155

Like? Yes (2) | No   

mohana lakshmi v   10 years AGO

by adding sequence no

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palasiva   10 years AGO

t(n)=n(n^2+n+1)
t(5)=5(5^2+5+1)=5(25+5+1)=155

Like? Yes (2) | No   

akash kumar   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (2) | No   

jitendra varshney   10 years AGO

t(n)=n^3 + n^2 + n
we have to find 5th term so put the value n=5
t(5)=5^3+5^2+5
=125+50+5
=155 ans..

Like? Yes (2) | No   

NAVEEN KUMAR   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n
Put n=5
T(5)=125+25+5

Like? Yes (3) | No (1)  

chandra mondal   10 years AGO

3*1=3
(3+4)*2=14
13*3=39
21*4=84
31*5=155

Like? Yes (2) | No   

Soham Chakraborty   10 years AGO

(3*1),(7*2), (13*3),(21*4),(31*5)

Like? Yes (2) | No   

chandanailupuru   10 years AGO

3*1,7*2,13*3,21*4=3,14,39,84
7-3=4,13-7=6,21-13=8
Next number is 21+10=31
31*5=155
Ans 155

Like? Yes (2) | No   

Partha Pratim Mahata   10 years AGO

3=1*[(1+1)^2-1]
14=2*[(2+1)^2-2]
39=3*[(3+1)^2-3]
...
..
155=5*[(5+1)^2-5]=5*31

Like? Yes (2) | No   

santu ghosh   10 years AGO

u(n)=n^3+n^2+n,n=1,2,3,4,5.....
u(1)=1^3+1^2+1=3
u(2)=2^3+2^2+2=14
.........................................
.....................................
u(5)=5^3+5^2+5=155

Like? Yes (2) | No   

madhurima   10 years AGO

1st no is:(1^2+2)*1
2nd no is:(2^2+3)*2
3rd no is:(3^2+4)*3
4th no is:(4^2+5)*4
so the 5th no will be:(5^2+6)*5=155
so,option 9 is correct

Like? Yes (2) | No   

D.HARINATH REDDY   10 years AGO

1^3+2=3
2^3+6=14
3^3+12=39
4^3+20=84
5^3+30=155

Like? Yes (2) | No   

G.Jaya Krishna   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
155=5*31

Like? Yes (2) | No   

Harsh Poddar   10 years AGO

1^3+1^2+1=3
2^3+2^2+2=14
3^3+3^2+3=39
4^3+4^2+4=84
next number=
5^3+5^2+5=155

Like? Yes (2) | No   

Chirag Ranjan Lahiri   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (2) | No   

deekshith   10 years AGO

it is the answer coz check out solution contact me @deekshith.nani@gmail.com

Like? Yes (2) | No   

Chiranjit Agarwal   10 years AGO

(3*1) (7*2) (13*3) (21*4) (31*5)
the difference between 3,7,13,21,31 is of the order 4,6,8,10 respectively

Like? Yes (3) | No (1)  

suman kumar   10 years AGO

differences are 11,25,45;
differences b\w them are 14,20 so next time it will be26.....
so adding it to 45 we get 71
since 71 is difference thus adding it to 84 we get 155.

Like? Yes (2) | No   

ravi shah   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No   

vidhijain   10 years AGO

Differnce of differnce is in a.p.

Like? Yes (1) | No   

vishva shikha   10 years AGO

3*1=3
7*2=14 (7-3)=4
13*3=39 (13-7)=6
21*4=84 (21-13)=8
21+10=31
31*5=155
hence answer=155

Like? Yes (1) | No   

Janaga mageshkumar.v   10 years AGO

diff bt two values 14-3,39-14,84-39 are in the order of 11 25 45 and diff bt these numbers are 14 20 so then diff as 26. so add 26+45=71 then 71+84=155

Like? Yes (1) | No   

prabhat kumar   10 years AGO

1*3, 2*7, 3*13, 4*21, 5*31

Like? Yes (1) | No   

potani manikumar   10 years AGO

3= 1*(1+1)^2 -1
.
.
.
5*(5+1)^2 -5

Like? Yes (1) | No   

Tark Patel   10 years AGO

1 ( 1^3 + 1^2 + 1) ,
14 ( 2^3 + 2^2 + 2) ,
39 ( 3^3 + 3^2 + 3) ,
84 ( 4^3 + 4^2 + 4) ,

so next number is
5^3 + 5^2 + 5 = 155

Like? Yes (1) | No   

vikash kumar   10 years AGO

3=1^3+1^2+1
14=2^3+2^2+2
39=3^3+3^2+3
84=4^3+4^2+4
155=5^3+5^2+5

Like? Yes (1) | No   

ADARSH GAURAV   10 years AGO

3=1^3+1^2+1^1
14=2^3+2^2+2^1
39=3^3+3^2+3^1
84=4^3+4^2+4^1
155=5^3+5^2+5^1

Like? Yes (1) | No   

Abhishek Mitra   10 years AGO

I think. It’s a sequence, the formula of which is for any number ‘n’, (n-1)3+(n-1)2+(n-1).

Like? Yes (1) | No   

Gaurav shah   10 years AGO

3*1,7*2,13*3,21*4,31*5

Like? Yes (1) | No   

SUNITEE GARIMA   10 years AGO

1^3+2=3
2^3+6=14
3^+12=39
4^3+20=84
5^3+10=155
so the next number is 155

Like? Yes (1) | No   

akshay   10 years AGO

5*31

Like? Yes (1) | No   

abhishek   10 years AGO

(1^3)+(1*2)=3
(2^3)+(2*3)=14
(3^3)+(3*4)=39
(4^3)+(4*5)=84
then
(5^3)+(5*6)=155

Like? Yes (1) | No   

Aravinth kumar   10 years AGO

a(n) = n^3 + n^2 + n
a(1)=1^3 + 1^2 + 1 = 3
... a(5)=5^3 + 5^2 +5 = 155

Like? Yes (1) | No   

sruthi   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No   

NIRUPOM GHOSH   10 years AGO

next no 155

Like? Yes (1) | No   

Enukonda Ramakrishna   10 years AGO

3, 14, 39, 84, x=155
11 25 45 71
14 20 26
6 6

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ABHIJIT DUTTA   10 years AGO

1*3 = 3
2*7 = 14
3 *13 = 39
4*21 = 84
similarly, 5*31 = 155

option 9)

Like? Yes (1) | No   

Tapobrata Banerjee   10 years AGO

we can get it using back triangle method.i.e. after finding the differences of given numbers we can get the series as 11,25,45 and the unknown no...then again 14,20 and the no..like this....finally 6 will be the common difference and by back calculation we can get the no. 155

Like? Yes (1) | No   

Ravi Bhalani   10 years AGO

5^3 + 5^2 +5=155

Like? Yes (1) | No   

HARDIP BHALANI   10 years AGO

5^3 + 5^2 +5=155

Like? Yes (1) | No   

Harshad    10 years AGO

5^3 + 5^2 +5=155

Like? Yes (1) | No   

Varunraj Reddy   10 years AGO

n^3+n^2+n

Like? Yes (1) | No   

Ankit Raj   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No   

shaurya neel   10 years AGO

n

Like? Yes (1) | No   

Sachin Bhakt   10 years AGO

(n-1)3+(n-1)2+(n-1).
n= nth number of term

Like? Yes (1) | No   

Sai Indu   10 years AGO

3*1=3
7*2=14
13*3=39
21*4=84
31*5=155
because the difference between 3 and 7 is 4,7 and 13 is 6,13 and 21 is 8,21 and 31 is 10...
as the difference order is 4,6,8,10.....so the answer is 155

Like? Yes (1) | No   

Rahul shaw   10 years AGO

By finding common differences...

Like? Yes (1) | No   

jagriti goomer   10 years AGO

general term of series:
(n)^3 + (n)^2 + (n)^1

putting n=5 for 5th term v get:
125+25+5=155

Like? Yes (1) | No   

latika   10 years AGO

3*1+0, 3*3+5,3*7+18,3*13+45,3*21+92

Like? Yes (1) | No   

shruthi   10 years AGO

(3+4)*2=14
(7+6)*3=39
(13+8)*4=84
(21+10)*5=155

Like? Yes (1) | No   

surya   10 years AGO

n^3+n^2+n

Like? Yes (1) | No   

gouthamdaneti   10 years AGO

n((n+1)^2-n) n denotes the position of number

Like? Yes (1) | No   

Hari Nani   10 years AGO

n((n+1)^2-n) n denotes the position of number

Like? Yes (1) | No   

Jasmita Parmar   10 years AGO

3 = 3 * 1
14 = 7 * 2
39= 13 * 3
84= 21 * 4
155= 31 * 5

Like? Yes (1) | No   

RAKESH KUMAR RANJAN   10 years AGO

1^3+2=3
2^3+6=14
3^3+12=39
4^3+20=84
5^3+30=155

Like? Yes (1) | No   

Niraj   10 years AGO

T(n) = n^3 + n^2 + n
So T(5) = 5(25+5+1) = 5*31 = 155

Like? Yes (1) | No   

ravi chandra singh   10 years AGO

3=1*3
14=2*7
39=3*13
84=4*21
155=5*31

Like? Yes (1) | No   

Shubhangi   10 years AGO

number-2*2+11

Like? Yes (1) | No   

IMRAN KHAN   10 years AGO

3*1+0
3*3+2+3
3*7+4+14

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Ajay Kumar Gupta   10 years AGO

(2-1)^3+(2-1^)2+(2-1)= 1 + 1 + 1 = 3
(3-1)^3+(3-1)^2+(3-1) = 8 + 4 + 2 = 14
continuing
(5-1)^3+(5-1)^2+(5-1)=64+16+4=84
then-(6-1)^3+(6-1)^2+(6-1)=125+25+5=155 ans.

Like? Yes (1) | No   

pavankumar   10 years AGO

choose ‘n’= (n-1)3+(n-1)2+(n-1).
choose n as 1
choose n as 2
choose n as 7

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RAJEEV KUMAR   10 years AGO

5*31=155

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Satya Karteek   10 years AGO

3= 1^3+1^2+1
14=2^3+2^2+2
similarly,
5^3+5^2+5=125+25+5=155

Like? Yes (1) | No   

vandna   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
155=5*31

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Ramu Giri   10 years AGO

n^3+n^2+n

where 0

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shree   10 years AGO

155.......

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shiv pujan   10 years AGO

sequence is n^3+n^2+n
putting n=5
ans is 155

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prakash kumar   10 years AGO

1^3+2=3,2^3+6=14,3^3+12=39,4^3+20=84,
5^3+30=155

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KRISHAN MURARI   10 years AGO

use n^3+n^2+n
put n=1,2,3........

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Mahak   10 years AGO

2*1+1 ,3*4+2 ,4*9+3, 5*16+4,
i.e.
2*(1)power2+1
3*(2)power2+2
4*(3)power2+3
5*(4)power2+4
so next number will be
6*(5)power2+5=155

Like? Yes (1) | No   

sudipta maji   10 years AGO

n3+n2+n=no
5^3+5^2+5=155

Like? Yes (1) | No (1)  

mantena phanindra v varma   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
155=5*31

Like? Yes (1) | No (1)  

Pushpendra singh chundawat   10 years AGO

(1^3+1^2+1)=3
(2^3+2^2+2)=14
(3^3+3^2+3)=39
(4^3+4^2+4)=84
(5^3+5^2+5)=155

opt-9

Like? Yes (1) | No (1)  

chandan jha   10 years AGO

(3+4)*2=14
(7+6)*3 =39
(13+8)*4=84
(21+10)*5=155..

Like? Yes (1) | No (1)  

ANKITA prasad   10 years AGO

1^3+2=3
2^3+6=14
3^3+12=39
4^3+20=84
5^3+30=155.
the number being added are increasing in multiples of 2.

Like? Yes (1) | No (1)  

Inline Geek   10 years AGO

Look at the series:

(1+1) * (1+1) - 1 = 3
(2+1) * (2+3) - 1 =14
(3+1) * (5+5) - 1 =39
(4+1) * (10+7) - 1 =84

so

(5+1) * (17+9) - 1 =155
is the answer

Like? Yes (1) | No (1)  

KRISHNA CHANDRA UPADHYAY   10 years AGO

Use expression n^3+n^2+n
and put n=1,2,3,4 and 5
after putting n=5 in above expression we get 155
HENCE.OPTION (9) IS CORRECT.

Like? Yes (1) | No (1)  

Dipak Singh   10 years AGO

webbed A.P
by caculating it comes 155

Like? Yes (1) | No (1)  

manish kumar singh   10 years AGO

1*3 2*7 3*13 4*21 5*31 6*43........

Like? Yes (1) | No (1)  

yatish kumar sharma   10 years AGO

12*12+11

Like? Yes (1) | No (1)  

Siva Prakash   10 years AGO

T(n) = n * (n^2 + n + 1)

1.) T(1) = 1 * (1^2 + 1 + 1) = 1 * 3 = 3

2.) T(2) = 2 * (2^2 + 2 + 1) = 2 * 7 = 14

3.) T(3) = 3 * (3^2 + 3 + 1) = 3 * 13 = 39

4.) T(4) = 4 * (4^2 + 4 + 1) = 4 * 21 = 84

5.) T(5) = 5 * (5^2 + 5 + 1) = 5 * 31 =155

Like? Yes (1) | No (1)  

Yogesh   10 years AGO

3*1=3
7*2=14
13*3=39
21*4=84
31*5=155

Like? Yes (1) | No (1)  

vishal   10 years AGO

1^3 +2 , 2^3 +6 , 3^3+12, 4^3 + 20 so ans is 5^3 +30

Like? Yes (1) | No (1)  

kajol nigam   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

13=(3^2+3+1)
21= (4^2 + 4 + 1)

for 5th = (25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (3) | No (3)  

Mogilipuri.Harish   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
here n=5 then
T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (1)  

rahul anand   10 years AGO

5*31

Like? Yes (1) | No (1)  

vinusha reddy   10 years AGO

1+1^2+1^3=3
2+2^2+2^3=14
i.e, t(n)=n+n^2+n^3
5+5^2+5^3=5+25+125=155

Like? Yes (1) | No (1)  

suman kumar   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (1)  

Mrityunjay   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

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Nitin Srivastava   10 years AGO

let x be next number
take the consecutive difference
step 1: 11 25 45 x-84
step 2: 14 20 x-129
step 3: 6 x-149
let next step difference be 0
x-149-6=0
x=155

Like? Yes (1) | No (1)  

Amritha   10 years AGO

3=1*3
14=2*7
34=3*13
84=4*21
the difference between 3 and 7 is 4, 7 and 13 is 6, 13 and 21 is 8 hence next difference must be 10.
So 5 * (21+10)= 155

Like? Yes (1) | No (1)  

raju manpreet singh   10 years AGO

2*1+1=3
4*3+2=14
6*6+3=39

8*10+4=84
10*15+5=155

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Sneha   10 years AGO

3=1*3
14=2*7
39=3*13
84=4*21
155=5*31
So option 9

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Dr.Dipin Singh   10 years AGO

nth term = n^3 + n^2 + n
5th term = 5^3+5^2+5= 125+25+5=155

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TINKU MITTAL   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (1)  

Manikumar R   10 years AGO

n^3 + n^2 + n
1^3 + 1^2 + 1=3
2^3+2^2+2=14
.....5^3+5^2+5=155

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selvakumar   10 years AGO

answer option 9)155

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chirag shrimali   10 years AGO

N^3+N^2+N

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Ramya   10 years AGO

1^3+1^2+1=3
2^3+2^2+2=14
.
.
5^3+5^2+5=155

Like? Yes (1) | No (1)  

vaishnavi   10 years AGO

3=1*3
14=2*7
39=3*13
84=4*21
so, next number will be 5*31=155

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S.SHAFI   10 years AGO

Formula = n^3 + n^2 + n

n=1 --> 1 + 1 + 1 = 3
n=2 --> 8 + 4 + 2 = 14
n=3--> 27+9+ 3 = 39
n=4--> 64 + 16 + 4 = 84
n=5 --> 125 + 25 + 5 = 155..

so ans is 155..

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sudhakar   10 years AGO

The answer is 155

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anandhakumar   10 years AGO

answer is 155
option 9

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Kushagra Tiwari   10 years AGO

we can see that
3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

we know that

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

so

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

answer = option 9, 155

Like? Yes (2) | No (5)  

Samrat Adhikari   10 years AGO

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (4)  

Amrita priyadarshi   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (4)  

Jeevanantham.V.M   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

Like? Yes (1) | No (4)  

Surit Datta   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155
thats it...

Like? Yes (1) | No (4)  

abhik k. dey   10 years AGO

sequence is 3, 14, 39, 84
now
3=1^3+2=1+2
14=2^3+6=8+6
39=3^3+12=27+12
84=4^3+20=64+20
now next number will be 5^3+30
as 2,6,12,20 in a common series where
6=2+4
12=6+6
20=12+8
so next will be 20+10=30
so for the above series as 3, 14, 39, 84 the next number will be 5^3+30=125+30=155

Like? Yes (1) | No (5)  

Abhishek Vishwakarma   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21

T(n) = n[(n+1)^2 - n]
= n(n^2 + 2n + 1 - n)
= n(n^2 + n + 1)
= n^3 + n^2 + n

T(5) = 5(25 + 5 + 1)
= 5 * 31
= 155

ans=155

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RAKESH   10 years AGO

n^3+n^2+n

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Sarabeshwar   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
155=5*31 Here consider 155 as term

while, Term(n) = n [ ( (n+1)to the power of 2 )- n ]
so Term (5) = 5[ (6)square - 5 ]
= 5 * 31 =155

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Devendra Marghade   10 years AGO

n+n^2+n^3
1+1^2+1^3= 1+1+1= 3
2+2^2+2^3=2+4+8= 14
3+3^2+3^3=3+9+27=39
4+4^2+4^3=4+16+64=84
5+5^2+5^3=5+25+125=155

Like? Yes (1) | No (5)  

KAUSHAL KUMAR   10 years AGO

14-3=11 25-11=14 20-14=6
39-14=25 45-25=20 next no=20+6=26
84-39=45 next no.=45+26=71
next no=84+71=155. option(9).

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kaushal kishor   10 years AGO

difference between 3,14,39,84 is 11,25 and 45 respectively.
and the difference between their diferences are 14,20,
which means 6, so next will be 26 and so 45+26=71
and so the next sequence will be 84+71=155

Like? Yes (2) | No (6)  

Arijit Acharyya   10 years AGO

1*3=3
2*7=14
3*13=39
4*21=84
5*31=155
next number=n*x
where n=1,2,3,4,5....
x=3,7,13,21,31,43... (difference is increasing by 2 starting from 4 )

Like? Yes (1) | No (5)  

sabir ahmd   10 years AGO

3 = 1 * 3
14 = 2 * 7
39 = 3 * 13
84 = 4 * 21
hence next no will be =5*31=155

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Aakash Jain   10 years AGO

1*3=3 leave next odd no-5
2*7=14 leave next 2 odd no's- 9,11
3*13=39 leave next 3 odd no's- 15,17,19
4*21=84 leave next 4 odd no's- 23,25,27,29
5*31=155

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Md.Sultan Khichi   10 years AGO

The series is like that
3*1=3
After this 5 is prime no. but leave it
7*2=14
Agn 11 is the next prime no. So leave it too.
13*3=39
But from now two prime no. Will be leave.
21*4=84.
Next two prime no. 23,29 will be leave so next is
31*5=155.
Ans. 155

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Rahul Sah   10 years AGO

3 14 39 84
11 25 45
14 20
6

anwer will be 6+20+45+84=155

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Rashmi bharti   10 years AGO

31*5=155

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AMIT SINGH   10 years AGO

difference between two nos. are=11,25,45
difference between the diff. are=14,20
so, next no.=20+6=26
so, next diff.=45+26=71
therefore, next sequence no.=84+71=155.

Like? Yes (2) | No (7)