What is the value of 0-1+2-3+4-5+6-7+...........+16-17+18-19+20.
Answer
625.8k+ views
Hint: In this question it is given that we have to find the value of the given series,
0-1+2-3+4-5+6-7+...........+16-17+18-19+20.
So to find the solution we need to know that if any series: a+(a+d)+(a+2d)+.......
is in A.P(Arithmetic Progression) then the summation of first n terms is,
$$S=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$.......(1)
Where a and d are the first term and common difference respectively,
So in order to get the solution we have to rearrange the given series.
Complete step-by-step solution:
So the given series,
0-1+2-3+4-5+6-7+...........+16-17+18-19+20
Now we are going to rearrange the positions by taking all the negative terms in one side,
So we can write,
0-1+2-3+4-5+6-7+...........+16-17+18-19+20
=0+2+4+6+8+10+12+14+16+18+20-1-3-5-7-9-11-13-15-17-19
=(2+4+6+8+10+12+14+16+18+20)-(1+3+5+7+9+11+13+15+17+19)
=$$S_{1}-S_{2}$$(say)
Where,
$$S_{1}=2+4+6+8+10+\ldots +20$$
$$S_{2}=1+3+5+7+9+11+\ldots +19$$
Now for $$S_{1}$$ the first term is a=2 and common difference, d=2-0=2 and total number of elements are n=10,
Therefore by formula (1) we can write,
$$S_{1}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$
$$=\dfrac{10}{2} \left( 2\times 2+\left( 10-1\right) \times 2\right) $$
$$=5\times \left( 4+9\times 2\right) $$
$$=5\times 22$$
$$=110$$
Now similarly for $$S_{2}$$ the first term is a=1 and common difference, d=3-1=2 and total number of elements are n=10,
Therefore by formula (1) we can write,
$$S_{2}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$
$$=\dfrac{10}{2} \left( 2\times 1+\left( 10-1\right) \times 2\right) $$
$$=5\times \left( 2+9\times 2\right) $$
$$=5\times \left( 2+18\right) $$
$$=5\times 20$$
$$=100$$
So we can write,
$$S_{1}-S_{2}=110-100=10$$
Therefore, the value of 0-1+2-3+4-5+6-7+...........+16-17+18-19+20 is 10.
Note: While solving any series related problem you need to first check whether the given series follows any order or not (A.P, G.P or H.P), if not then you have to make it by doing rearrangement like we did while solving the above problem. Where after rearranging we get two series which are in G.P.
0-1+2-3+4-5+6-7+...........+16-17+18-19+20.
So to find the solution we need to know that if any series: a+(a+d)+(a+2d)+.......
is in A.P(Arithmetic Progression) then the summation of first n terms is,
$$S=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$.......(1)
Where a and d are the first term and common difference respectively,
So in order to get the solution we have to rearrange the given series.
Complete step-by-step solution:
So the given series,
0-1+2-3+4-5+6-7+...........+16-17+18-19+20
Now we are going to rearrange the positions by taking all the negative terms in one side,
So we can write,
0-1+2-3+4-5+6-7+...........+16-17+18-19+20
=0+2+4+6+8+10+12+14+16+18+20-1-3-5-7-9-11-13-15-17-19
=(2+4+6+8+10+12+14+16+18+20)-(1+3+5+7+9+11+13+15+17+19)
=$$S_{1}-S_{2}$$(say)
Where,
$$S_{1}=2+4+6+8+10+\ldots +20$$
$$S_{2}=1+3+5+7+9+11+\ldots +19$$
Now for $$S_{1}$$ the first term is a=2 and common difference, d=2-0=2 and total number of elements are n=10,
Therefore by formula (1) we can write,
$$S_{1}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$
$$=\dfrac{10}{2} \left( 2\times 2+\left( 10-1\right) \times 2\right) $$
$$=5\times \left( 4+9\times 2\right) $$
$$=5\times 22$$
$$=110$$
Now similarly for $$S_{2}$$ the first term is a=1 and common difference, d=3-1=2 and total number of elements are n=10,
Therefore by formula (1) we can write,
$$S_{2}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$
$$=\dfrac{10}{2} \left( 2\times 1+\left( 10-1\right) \times 2\right) $$
$$=5\times \left( 2+9\times 2\right) $$
$$=5\times \left( 2+18\right) $$
$$=5\times 20$$
$$=100$$
So we can write,
$$S_{1}-S_{2}=110-100=10$$
Therefore, the value of 0-1+2-3+4-5+6-7+...........+16-17+18-19+20 is 10.
Note: While solving any series related problem you need to first check whether the given series follows any order or not (A.P, G.P or H.P), if not then you have to make it by doing rearrangement like we did while solving the above problem. Where after rearranging we get two series which are in G.P.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

