If x, (x^3 + 1) and x^4 are in arithmetic progression, then possible sum of the 3 terms of arithmetic progression will be (Given taht x is real number)

M4maths Previous Puzzles

16 Jul 2015 Arithmetic Progression If x, (x^3 + 1) and x^4 are in arithmetic progression, then possible sum of the 3 terms of arithmetic progression will be (Given taht x is real number): OPtion
1) 27, 0
2) 25, 1
3) 20, -1
4) 27, -1
5) 25, 0
6) 20, 1
7) 27, 1
8) 25, -1
9) 20, 0
10)18, 2; Correct Option: 1 Best Solution (45)

SAYAN CHATTERJEE 9 years AGO

x, x^3+1 & x^4 are in AP
hence, 2x^3+2= x^4+x
=> x^4-2x^3+x-2=0
this has two factors x=-1 & x=2
so, the summation of the terms are 0 & 27 (1)

Like? Yes (1) | No

ramakrishnateja 9 years AGO

given that x,x^3+1,x^4 are in a.p.hence x^3+1-x=x^4-(X^3+1)
x^4-2x^3+x=2. so x=2 and x=-1. sum=27,0

Like? Yes (1) | No

sudiptaa 9 years AGO

x=2
series : 2, 9,16
sum=27
x= -1
series : -1,0,1
sum=0

Like? Yes (1) | No

RAMPALA JYOTHI 9 years AGO

x^3+1-x=x^4-x^3-1;
x^4-2x^3+x-2=0;
there fore x= -1 & 2 (x is real);
sum of three terms:
if x=-1
x+x^3+1+x^4=0;
if x=2
x+x^3+1+x^4=27;
so ans: 27,0...................

Like? Yes (1) | No

himanshu 9 years AGO

A.P=a,a+d,a+2d
given x,x^3+1,x^4
x=a
a+d=x^3+1
a+2d=x^4
put the value of d
we get x^4 -2x^3+x-2=0
put x=2 and x= -1 we get 0
therefore add x+x^3+1+x^4 we get 27,0

Like? Yes (1) | No

UJJWAL KANT 9 years AGO

since x, x^3+1 and x^4 are in AP
x^3+1-x=x^4-(x^3+1)
from here we get two values of x i.e
x=2 and x=-1
putting these value we get
for x=2 AP will be 2, 9, 16 and sum=27
lly for x=-1 AP will be -1, 0, 1 and sum will be 0

Like? Yes (1) | No

sudhanshu ranjan 9 years AGO

on solving the equation x^4 - 2x^3 + x - 2 =0 we get x=2 and x= -1

Like? Yes (1) | No

Balaselvaraj M 9 years AGO

they tell us given series is a arithmatic progression and x is an integer.so we take 2 it satisfies the given data.

Like? Yes (1) | No

Ananya Singh 9 years AGO

(x+x^4)/2=x^3 +1
x=(2,-1)
sum=n/2(2a+(n-1)d)
sum=27,0

Like? Yes (1) | No

Deeksha Joshi 9 years AGO

since 3 terms are in ap so they hv same diffence .
therefore
x3+1-x=x4-x3-1
x4-2x3+x-2=0
(x+1)(x-2)(x2-x+1)=0
since x is real so x=-1,2

so
two sequences will b= -1,0,1
and 2,9,16
sum of the ap is= n*(a+l)/2
so sum of 2 aps is= 27 and 0.

Like? Yes (1) | No

aashish kumar 9 years AGO

assuming x=2.
x+(x^3+1)+x^4
=2+(2^3+1)+2^4
=2+9+16
=27

Like? Yes (1) | No

Soma Naresh 9 years AGO

difference
x^3+1-x=x^4-(x^3+1)
2x^3+2-x^4-x=0
2 (x^3+1)-x (x^3+1)=0
(2-x)(x^3+1)=0
x= 2,-1
x=2
sum=27
x=-1
sum=0

Like? Yes (1) | No (1)

Deepak kushwaha 9 years AGO

here we have AP
x,x^3+1,x^4
we know that in AP
mean of first and third digit is equal to second digit
(x+x^4)/2=(x^3+1)
x^4+x=2x^3+2
x^4-2x^3+x-2=0
x^3(x-2)+1(x-2)=0
(x^3+1)(x-2)=0
on solving we get
x=2,-1
on putting these value we get
AP1=2,9,16
AP2=-1,0,1
sum of AP1=27
sum of AP2=0
possible sum is 27,0 (option 1)

Like? Yes (3) | No (3)

NIRBHAY KUMAR 9 years AGO

The progression satisfies on x=2
now progression will be 2,9,16
And the sum of this progression is 27.

Like? Yes (1) | No (1)

SOMA NARESH 9 years AGO

difference between terms is
x^3 +1 - x = x^4 -(x^3+1)
x^3 +1 - x = x^4 -x^3-1
2x^3 +2 = x^4+x
2(x^3+1)-x(x^3+1)=0
x = 2 x = -1
Sum when x=2 is 2+ 2^3+1+ 2^4 = 2+9+16 = 27
Sum when x=2 is -1+(-1+1)+1 = 0

Like? Yes (1) | No (1)

anand vijay 9 years AGO

x,x^3+1,x^4 are in a.p.
x^3+1-x=x^4-x^3+1.....(i)
x(x-1)(x^2-x-1)=2
x=-1,2,3
on putting the value of x in equation (i) only x=-1,2
satisfy the equation
hence,
sum of a.p=x+x^3+1+x^4
=27 (x=2),0(x=-1)
option (i) is the right answer

Like? Yes (1) | No (1)

ROHIT KUMAR 9 years AGO

For arithmetic progression x satisfies two value
I.e x= -1 and 2
then sum of 3 terms will be 0 and 27

Like? Yes (2) | No (2)

pooja jha 9 years AGO

x,x^3+1 and x^4 are in ap.
so x^3+1-x=x^4-x^3-1.
x^4-2x^3+x-2=0
therefore x=2 and x=-1 are the two solution.
therefore sum of three numbers are 27 and 0 on putting x=2 and x=-1 respectively in the equation

Like? Yes (1) | No (1)

Dinesh kumar 9 years AGO

x, (x^3 + 1) and x^4 are in arithmetic progression
Therefore (X^3+1)-X=X^4-(X^3+1)
X^4-2X^3+X-2=0
X^3(X-2)+1(X-2)=0
(X^3+1)(X-2)=0
X^3=-1 X=2
X=-1 X=2
sum of the 3 terms of arithmetic progression will be
X+(X^3+1)+X^4
sub X=-1
-1+(-1+1)+(-1)^4=0
sub X=2
2+(8+1)+2^4=11+16=27
Therefore ans is 27,0

Like? Yes (1) | No (1)

karthiga 9 years AGO

when x=-1;
A.P = -1 , 0 , 1;
Sum is 0;
when x= 2;
A.P = 2 , 9 , 16;
Sum is 27;

Like? Yes (1) | No (1)

Kethineni Vinodh 9 years AGO

2b=a+c
then (x^3+1)(x-2)=0
=>x = 2 or -1
sum= 27 or 0

Like? Yes (1) | No (1)

SURAJ NIRALA 9 years AGO

2nd term - 1st=3rd- 2nd term
equivalent to x^4-2x^3+x-2=0

x=2. require number is 2, 9, 16
sum 27

Like? Yes (1) | No (1)

Ashu Kumar 9 years AGO

X,X^3+1AND X^4 ARE IN AP
SO 2(X^3+1)=X+X^4
SO X=-1,2
FOR X=-1 TERMS ARE -1,0AND 1
SO SUM=0
FOR X=2 TERMS ARE 2,9AND 16
SO SUM=27
SUM=27,0

Like? Yes (2) | No (2)

Amit Roy 9 years AGO

as it is an AP series so x^3+1-x=x^4-(x^3+1)
there after solving this equality we found x=2 and -1.
putting the value of x(=2) we found series is 2,9,16 i.e sum=27 & x(=-1) we found -1,0,1 i.e sum=0.

Like? Yes (2) | No (2)

Vibhu Banka 9 years AGO

since the given terms are in A.P.,
so, 2(x^3+1)=x + x^4
this gives, x=2,-1
therefore possible sum= 27 or 0

Like? Yes (1) | No (1)

Dinni 9 years AGO

X^4-2X^3+X-2=0.
X=2,-1.
X=2, then sum=2+9+16=27.
X=-1, then sum=-1+0+1=0.
So, 27 &0 are the answers.

Like? Yes (1) | No (1)

shivjee yadav 9 years AGO

A/C to AP (X^3+1)-X=X^4-(X^3+1)

This progression satisfied on two values of x=-1, 2
SO,Answer will be 27,0

Like? Yes (1) | No (1)

Abhishek 9 years AGO

Since terms are in A.P. (x^4 + x)/2 = x^3+1
on simplifying it, we get x^4 - 2 x^3+x-2 = 0
two of its roots are 2,-1 other two roots are imaginary.

now sum of terms using 2 as a root 2+2^3+1+2^4=27.
using -1 as root (-1)+(-1)^3+1+(-1)^4 = -1+(-1)+1+(1) = 0
so sum of 3 terms are 27 , 0

Like? Yes (1) | No (1)

Rahul kumar 9 years AGO

since given series is in ap we can take common diff as
2nd -1st =3rd -2nd
so we get
x^3+1-x=x^4-x^3-1
solving it we get x=2,-1
put once x=2&x=-1
also Sn=n/2(a+l)
we will get 2 series
1) 2,9,4 whose sum is 27
2)-1,0,1 whose sum is 0

Like? Yes (1) | No (1)

Deepak 9 years AGO

Real roots of the equation are: x = 2 and x = -1,
Sum of three terms: x^4 + x^3 + x +1; 27, 0

Like? Yes (2) | No (2)

sai 9 years AGO

given that the 3 are in a.p and x is a real number..the number that satisfies the condition is 2.. then the sum is 2+9+16=27

Like? Yes (1) | No (1)

GITANJALI GOSWAMI 9 years AGO

as it is in Ap,differences between successive terms are equal.So
(x^3+1)-x=x^4-(x^3+1)
=>x^4-2x^3+x-2=0
=>x^3(x-2)+1(x-2)=0
=>x=2
or
x=-1

so summation will be
for x=2,2+(8+1)+16=27
for x=-1,-1+(-1+1)+(-1)^4=0

Like? Yes (1) | No (1)

vivek 9 years AGO

if x= -1 thn,
-1,0,1..are in ap.
so sum of three term of series will be..27,0.

Like? Yes (1) | No (1)

Pinku Burnwal 9 years AGO

in A.P
2(2nd term)=1st term+3rd term
2(x^3+1)=x(x^3+1)
which give x=3,-1
which give sum of three term=27, 0
option 1

Like? Yes (1) | No (1)

PRITI MISHRA 9 years AGO

since, x, (x^3 + 1) and x^4 are in arithmetic progression
2(x^3+1)=x+x^4
2(x^3+1)=x(x^3+1)
(x^3+1) (2-x)=0
x=2,1,1,-1
hence,
for x=2, we have
x+(x^3 + 1) +x^4=27
and for x=-1
sum will be 0

Like? Yes (1) | No (1)

Ajay Kumar Gupta 9 years AGO

the no. is x=2 and -1
the terms are 2,9,16
and -1,0,1
so the sum of three terms are 27,0

Like? Yes (1) | No (1)

Kirty 9 years AGO

by equating common difference
x^3+1-x=x^4-(x^3+1)
X^4+2*x^3+x-2=0
solving this we get as x=-1 or x=2
by putting x=2 the sum comes to be 27
and by putting x=-1 the sum comes to be 0

Like? Yes (1) | No (1)

novel_liker 9 years AGO

values of x are calculated as -1,-1,-1,2
arithmatic series : -1,0,-1 so sum is 0
arithmatic series : 2,9,16 so sum is 27
hence option 1 satisfies (27,0)

Like? Yes (1) | No (1)

shiva 9 years AGO

Given that X, X^3+1, X^4 are in A.P and also given that X is a real number
Lets assume X=2 , -1
If X=2 then the terms are 2 ,9 ,16
Here the common difference d=7
If X= -1 then the terms in A.P are -1 , 0, 1
Here the common difference d= -1
Hence the possible sum of three terms is 27 , 0

Like? Yes (1) | No (1)

Manish kumar 9 years AGO

(x^3+1)-x=x^4 -(x^3+1)
=>x^4-2x^3+x-2=0
aftr solving we get
x=2,-1(real no.)
thus sum=27,0

Like? Yes (1) | No (1)

SAURABH SINGH 9 years AGO

when we put value like x=-1 their sum becomes zero.
and if we put values like x=2 their sum becomes 27.

Like? Yes (1) | No (1)

rahul shaw 9 years AGO

as numbers are in Arithmetic Possession
so
diff between nos are same...
(x^3+1)-x=(x^4)-(x^3-1)
after solving it...we get x=2 ,-�?1
given: x is a real no. so x= 2 and 0
adding all terms we get 27,0

Like? Yes (1) | No (1)

saikumar 9 years AGO

sum of ap=b=(a+c)/2
=x^3+1=(x+X^4)
x=2,-1
sum of terms=n/2(a+l)
if x=2 by sub above eqn ans=27
if x=-1
ans=0
(27,0)

Like? Yes (1) | No (1)

Shiva Sharma 9 years AGO

In AP 2*b=a+c
so 2*(x^3+1) = x+x^4
x^4+x-2*x^3-2=0
x^3+1=0 and x-2=0
x=2 and x^3 = -1
so sum = 27 and 0

Like? Yes (2) | No (2)

Piyush Telang 9 years AGO

Since x, x^3+1 and x^4 are in AP
Then 2 * ( x^3 + 1 ) = x + x^4 => x^4 - 2x^3 + x = 2
=> x ( x^3 -2x^2 + 1 ) = 2
=> x = 2 and x^3 -2x^2 + 1 = 2
=> x^3 -2x^2 = 1 => x^2 ( x - 2 ) = 1
=> x^2 = 1 and x- 2 = 1
=> x = +1 , -1 and 3
Thus possible values of x are -1, +1, +2 and +3.
But for values +1 and +3 the x, x^3+1 and x^4 does not follow to be in AP i.e. ( 1, 2, 1 ) and ( 3, 28, 81 )
Thus Sum is either 27 or 0 for the series ( 2, 9, 16 ) and ( -1, 0, 1 ) respectively.

Like? Yes (1) | No (2)

Dimple

India

Punjab

Time: 00:01:33

Points

Maths Quotes

"Mathematics is one of the ways to get enlightened"

prashanth

Mathematical proofs, like diamonds, are hard and clear, and will be touched with nothing but strict reasoning.

John Locke

Placed User Comments

M4Math helped me a lot.

Vipul Chavan 4 years ago

Thanks m4 maths for helping to get placed in several companies.
I must recommend this website for placement preparations.

yash mittal 5 years ago

If x, (x^3 + 1) and x^4 are in arithmetic progression, then possible sum of the 3 terms of arithmeti

M4maths Previous Puzzles

16 Jul 2015 Arithmetic Progression