What is the sixth term of following series: 5+7i, 7+14i, 11+55i, 19+110i, 35+121i, ....

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30 May 2017 Series

What is the sixth term of following series:

5+7i, 7+14i, 11+55i, 19+110i, 35+121i, ....

OPtion

1) 56 + 256i
2) 67 + 222i
3) 67 + 242i
4) 68 + 226i
5) 67 + 227i
6) 67 + 221i
7) 56 + 227i
8) 56 + 243i
9) 68 + 221i
10) None of these

Solution

The complex number is formed by two series:
Real part series is:
5, 7, 11, 19, 35, 67
nth term is 2^n+3 = 2^6+3 = 67

Imaginary part:
7, 14, 55, 110, 121, 242
Sum of previous number and its reverse = 7+7 = 14
Similarly 14+41 = 55
55+55 = 110
110+011 = 121
121+121 = 242

Correct Option: 3 Best Solution (22)

A.P.Singh   7 years AGO

First part of term is obtained by adding 2,4,8,16,32 in previous term like 5+2=7, 7+4=11, 11+8=19, 19+16=35 and 35+32=67.
second part is obtained by adding previous term and reversed previous term like 14+41=55, 110+011=121 and hence 121+121=242.

so 6th term is 67+242i.

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G.S.Kantharao   7 years AGO

5+7i,7+14i,11+55i,19+110i,35+121i...
If we observe the terms first terms in the given are 5,7=(5+2*1),11=(7+2*2),19=(11+2*4),35=(19+2*8),Next one is 67=(35+2*16)
2nd terms are 7i,2*7i,55i,2*55i,121i,So the next one is 242i=2*121i
Hence the Sixth term is 67+242i.
Hence the correct option is 3.☝

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abhishek   7 years AGO

first increasing with power of 2 i.e. 2,4,8,16 then 32
another term is double of previous one i.e. 7 becomes 14
55 becomes 110 then 121 becomes 242. therefore answer is 67+242i.
option 3)

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Jahnavi   7 years AGO

for real part
5*2-3=7
7*2-3=11
11*2-3=19
19*2-3=35
hence , 35*2-3=67
for imaginary part,
7*2=14
55*2=110
hence, 121*2=242
so,the answer is 67 + 242 i
ie. option 3

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Sivaprakash​   7 years AGO

5+2=7
7+4=11
11+8=19
19+16=35
35+32=37
7*2=14
55*2=110
121*2=242

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Nimalan   7 years AGO

the difference between the real part is in the form 2 4 8 16 therefore next one is 32
in imaginary part
7+7 = 14
14+41=55
55+55=110
110+11=121
121+121=242

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Amarnath Kumar Upadhyay   7 years AGO

for first term
5+2^1=7
7+2^2=11
11+2^3=19
19+2^4=35
35 + 2^5=67

and for second term add reverse of the second term to itself
7+7=14
14+41=55
55+55=110
110+011=121
121+121=242
so sixth term is 67+242i

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vinamra   7 years AGO

take a combination of 1&2 ,3%4 ,and 5&6
check (1st and 2nd) terms ....2nd term iorta part is a multiple of 2 *( integer part of iorta) i,e 2*7=14
same concept for 3rd and 4th term
2*55=110
as well as
6th term are 2*(integer part of iorta)
I,e 2*121=242
...that's y option 3 is correct
good nd easy trick☺

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BHAVANA   7 years AGO

5+2=7 and 7+7=14
7+4=11 and 14+41=55
11+8=19 and 55+55=110
19+16=35 and 110+011=121
35+32=67 and 121+121=242

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Chinthalachervu.silpa   7 years AGO

5×2-3=7.

7×2-3=11.
:
:
:
35×2-3=67 &
starting:7
7×2=14
55×2=110
121×2=242
so 67+242i

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tumati koushik   7 years AGO

for abscissa we have to multiply previous abscissa by 2 and then subtract 3 from it. and for alternate mantissa we have to double previous one.

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Ramandeep Kaur   7 years AGO

Each term incremented by itself and i series is incremented by its reverse to get new term

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pradeep reddy   7 years AGO

5+7, 7+14, 11+55, 19+110, 35+121
Take first series no's:
5, 7 , 11, 19, 35,......

Diff of this no's are in the form of 2^1, 2^2, 2^3, 2^4, 2^5,..........
so based on the above info 35+2^5=67

Take second series:
7, 14, 55, 110, 121

Here just add the orginal+reverse the original no:
ex: 7 +7=14
14+41=55
55+55=110
110+011=121
121+121=242
.
.
So Six term series is: 67+242

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pravin singh   7 years AGO

2,4,8,16,32 for 1st term
7X2=14
55X2=110
121X2=242
67 + 242i ans

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G.krishnaveni   7 years AGO

5+7i , 7+14i ,11+55i ,19+110i ,35+121i
5,7,11,19,35---
67
7,14,55,110,121--
242
so 67+242i

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Nitish kesarwani   7 years AGO

taking two pair of each in which Ith pair increase by 2 times
( 5+7i, 7+14i) ,(11+55i,19+110i) ,(35+121i,,68+242i)
so
option 3

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ganesh chemanchula   7 years AGO

when you observe first term in the series,5,7,11,19,35,..
the difference between adjacent terms is getting doubled..so next term will be 32+35=67
And in second term, 7,14,55,100,121...numbers are getting doubled..so the next term will be 121*2=242
that is 67+242i..option 3.

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palak Jain   7 years AGO

5,7,11,19,35,.....difference of 2^1,2^2,2^3,2^4,...next2^5
So 35+2^5=67+242i

Complex no..7i,14i,55i,110i,121i....
7+7=14
55+55=110
121+121=242

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Sumanth koditala   7 years AGO

First terms sequence 5, 5+2=7, 7+4=11 ,11+8=19 ,19+16=35, 6th term=35+32=67

Second term Sequence 7, 7*2=14 ,
55 , 55*2=110,
121 , 6th term =121*2=242

Overall : 67+242i

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Mukund Manohar   7 years AGO

Numbers without i part - 5,7,11,19,35
5+(2)=7
7+(2*2)=7+4=11
11+(4*2)=11+8=19
19+(8*2)=19+16=35
35+(16*2)=35+32=67
Numbers with i part -
7,14,55,110,121
Adding number by reversing it's digits
7+7=14
14+41=55
55+55=110
110+011=121
121+121=242
Sixth Term = 67+242i

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Gokul Krishnan   7 years AGO

5+7i,7+14i,11+55i,19+110i,35+121i...
7i*2=14i
55i*2=110i
121i*2=242i
and
5+2=7
7+4=11
11+8=19
19+16=35
35+32=67
hence answer is 67+242i

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K CHANDRA SHEKAR   7 years AGO

5,7,11,19,35
2,2^2,2^3,2^4,2^5
35+32=67 this is first term
7i,14i,55i,110i,121i,
7+7=14
55+55=110
121+121=242
ans=67+242i

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