1+3+5+.find sum upto 20 terms.
Answers
In the above series 1–3+5–7+9 , I believe that the series can be broken into two Arithmetic progressions
Series 1 will be 1+5+9+…….. and
Series 2 will be -3–7-……..
By using the formula Sum of first n terms for Arithmetic progress
S=(n/2)[2a+(n-1)d] where “a” is the first number in the series and “d” is the common difference.
a, a+d , a+2d, a+3d….. in similar way
For series 1: 1,(1 +4) + (1+2(4)) + ….
5–1=9–5 (b-a=c-b) so the common difference d is “4” and a is 1
The sum of first 20 terms of this series is
Sum1 = (n/2)[2a+(n-1)d]
=(20/2)[2(1) + (20–1)4]
=10[2+(19)4]
=10[78]
=780
so for Series 1 the sum of first 20 terms is 780
For Series 2: -3,-7 if we assume that the series is -3,-7,-11…
the a=-3 and the common difference “d” is b-a=c-b i.e. (-7) -(-3) = (-11) - (-7)
-7 +3 = -11+7 = -4
so the series is -3 , -3+(-4), -3 +2(-4),….
Sum2 = (n/2)[2a+(n-1)d]
= (20/2)[2(-3)+(20–1)(-4)]
= 10[-6–76] = 10[-82]
=-820
adding the Sum1 and Sum2 would give us the result “-40” and this is assuming that both the series go up to 20 terms.