Sum of 1, 3, 6, 10, 15, 21....... upto 10 terms

M4maths Previous Puzzles

26 Aug 2010 Find the sum

Sum of 1, 3, 6, 10, 15, 21....... upto 10 terms

OPtion

1) 231
2) 230
3) 225
4) 345
5) 197
6) 195
7) 220
8) 224
9) 218
10) 204

Solution

1+3+6+10+15+21+.......
=1+(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)+.......
= 1*10+2*9+3*8+4*7+5*6+6*5+7*4+8*3+9*2+10*1
=2(10+18+24+28+30)
=220

Correct Option: 7 Best Solution (20)

Mayur   13 years AGO

Sum of 1, 3, 6, 10, 15, 21.......55=220

Like? Yes (2) | No   

Karthik Pandiyan   13 years AGO

Since Difference between first 2 nos is one and difference is increasing one more for consective numbers in series. hence 10 terms will be 1,3,6,10,15,21,28(21+7),36(28+8),45(36+9),55(45+10)
On adding those 10 terms =220

Like? Yes (1) | No   

Dinesh Babu   13 years AGO

The numbers are 1 3 6 10 15 21 27 34 42 51 and their sum is 220

Like? Yes (1) | No   

S. Singh   13 years AGO

Sum of 1, 3, 6, 10, 15, 21....... upto 10 terms is 220.

Like? Yes (1) | No   

Dr.Dipin Singh   13 years AGO

Sum of 1, 3, 6, 10, 15, 21....... upto 10 terms = 1*10 + 2*9+3*8+4*7+5*6+6*5+7*4+8*3+9*2+10*1 = (10+18+24+28+30)*2 = 2*110=220

Like? Yes (1) | No   

Shikha   13 years AGO

1+2+3+....+55= 220

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kewal   13 years AGO

1+3+6+....+45+55 = 220

Like? Yes (1) | No   

Vasu   13 years AGO

Sum of 1, 3, 6, 10, 15, 21....... upto 10 terms is 220.

Like? Yes (1) | No   

SUDESHNA SEN   13 years AGO

f(1) = (1^2+1)/2= 1
f(2) = (2^2+2)/2= 3
f(3) = (3^2+3)/2= 6
f(4) = (4^2+4)/2= 10
f(5) = (5^2+5)/2= 15
f(6) = (6^2+6)/2= 21
f(7) = (7^2+7)/2= 28
f(8) = (8^2+8)/2= 36
f(9) = (9^2+9)/2= 45
f(10) = (10^2+10)/2= 55
So, the sum is= 1+3+6+10+15+21+28+36+45+55= 220..... Option 7)

Like? Yes (1) | No   

Mamon Sengupta   13 years AGO

f(1) = (1^2+1)/2= 1
f(2) = (2^2+2)/2= 3
f(3) = (3^2+3)/2= 6
f(4) = (4^2+4)/2= 10
f(5) = (5^2+5)/2= 15
f(6) = (6^2+6)/2= 21
f(7) = (7^2+7)/2= 28
f(8) = (8^2+8)/2= 36
f(9) = (9^2+9)/2= 45
f(10) = (10^2+10)/2= 55
So, the sum is= 1+3+6+10+15+21+28+36+45+55= 220..... Option 7)

Like? Yes (1) | No   

Trishita Dasgupta   13 years AGO

f(1) = (1^2+1)/2= 1
f(2) = (2^2+2)/2= 3
f(3) = (3^2+3)/2= 6
f(4) = (4^2+4)/2= 10
f(5) = (5^2+5)/2= 15
f(6) = (6^2+6)/2= 21
f(7) = (7^2+7)/2= 28
f(8) = (8^2+8)/2= 36
f(9) = (9^2+9)/2= 45
f(10) = (10^2+10)/2= 55
So, the sum is= 1+3+6+10+15+21+28+36+45+55= 220..... Option 7)

Like? Yes (1) | No   

ARKA GANGULY   13 years AGO

f(1) = (1^2+1)/2= 1
f(2) = (2^2+2)/2= 3
f(3) = (3^2+3)/2= 6
f(4) = (4^2+4)/2= 10
f(5) = (5^2+5)/2= 15
f(6) = (6^2+6)/2= 21
f(7) = (7^2+7)/2= 28
f(8) = (8^2+8)/2= 36
f(9) = (9^2+9)/2= 45
f(10) = (10^2+10)/2= 55
So, the sum is= 1+3+6+10+15+21+28+36+45+55= 220..... Option 7)

Like? Yes (1) | No   

ASHOK ROY   13 years AGO

f(1) = (1^2+1)/2= 1
f(2) = (2^2+2)/2= 3
f(3) = (3^2+3)/2= 6
f(4) = (4^2+4)/2= 10
f(5) = (5^2+5)/2= 15
f(6) = (6^2+6)/2= 21
f(7) = (7^2+7)/2= 28
f(8) = (8^2+8)/2= 36
f(9) = (9^2+9)/2= 45
f(10) = (10^2+10)/2= 55
So, the sum is= 1+3+6+10+15+21+28+36+45+55= 220..... Option 7)

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Arun prasath   13 years AGO

1+3+6+10+15+21+28+36+45+55=220

Like? Yes (1) | No   

DEBALINA CHAKRABORTY (DUTTA)   13 years AGO

f(1) = (1^2+1)/2= 1
f(2) = (2^2+2)/2= 3
f(3) = (3^2+3)/2= 6
f(4) = (4^2+4)/2= 10
f(5) = (5^2+5)/2= 15
f(6) = (6^2+6)/2= 21
f(7) = (7^2+7)/2= 28
f(8) = (8^2+8)/2= 36
f(9) = (9^2+9)/2= 45
f(10) = (10^2+10)/2= 55
So, the sum is= 1+3+6+10+15+21+28+36+45+55= 220..... Option 7)

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karthickma   13 years AGO

Sum of Natural numbers in order...

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A J V   13 years AGO

1+3+6+10+15+21+28+36+45+55=220. Obtained...

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YadeeshKumar   13 years AGO

1+3+6+10+15+21+28+36+45+55=220

Like? Yes (1) | No   

senthilkumaran   13 years AGO

Sum of the Given numbers upto 10 terms is 220.

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balaji   13 years AGO

1+3+6+10+15+21+28+36+45+55=220

Like? Yes (1) | No