What is the next number of the following sequence 14, 31, 41, 58, 147, ......

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25 Jan 2016 Sequence

What is the next number of the following sequence
14, 31, 41, 58, 147, ......

OPtion

1) 224
2) 213
3) 352
4) 327
5) 182
6) 297
7) 172
8) 176
9) 251
10)203

Solution

14
14 + 1^2 + 4^2 = 14+1+16 = 31
31 + 3^2 + 1^2 = 31+9+1 = 41
41 + 4^2 + 1^2 = 41+16+1 = 58
58 + 5^2 + 8^2 = 58+25+64 = 147
147 + 1^2 + 4^2 + 7^2 = 147+1+16+49 = 213
Option 2)

Correct Option: 2 Best Solution (11)

GOURANGA SUNDAR MANDAL   8 years AGO

the 1st term of series is 14 then the 2nd term =14+{(1)^2+(4)^2}=17...thus the 3rd term , 4th term are produced..so the last term is {147+(1)^2+(4)^2+(7)^2}=213....so option 2 is correct answer

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Pranav Sathe   8 years AGO

the difference between 2 consecutive numbers in the square of individual numbers within a number..eg
1*1 + 4*4 + 14=31
3*3+1*1 +31=41...1*1+4*4+7*7+147=213

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RAKESH   8 years AGO

14+ (1^2+4^2) ------ 31



147+(1^2+4^2+7^2) ---------- 213

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Devendra Marghade   8 years AGO

14+(1^2+4^2)=31
31+(3^2+1^2)=41
41+(4^2+1^2)=58
58+(5^2+8^2)=147
147+(1^2+4^2+7^2)=213
Option 2)

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satish ojha   8 years AGO

147+1+4^2+7^2=213

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MRITUNJAY   8 years AGO

14+4^2+1^2=31
31+3^2+1^2=41
41+16+1=58
58+25+64=147
147+1+16+49=213
option 2

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PRABHAT KUMAR   8 years AGO

14
14+1+16=31
31+9+1=4e1
41+16+1=58
58+25+64=147
147+1+16+49=213
Option 2

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Ramgopal   8 years AGO

14
14+1^2+4^2=31
31+3^2+1^2=41
41+4^2+1^2=58
58+8^2+5^2=147
147+1^2+4^2+7^2=203

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G BHARATH REDDY   8 years AGO

14+17=31
31+10=41
41+17=58
58+89=147
147+66=213

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Surabhi Yadav   8 years AGO

14=1^2+4^2=17
14+17=31=3^2+1^2=10
31+10=41=4^2+1^2=17
41+17=58
Similarly
1^2+4^2+7^2=1+16+49=66+147
=213

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sachin sawant   8 years AGO

1^2+4^2=17
14+17=31
Similarly,3^2+1^2=10
31+10=41
Thus,for147,(1^2+4^2+7^2=66)
147+66=213
Option2

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