find the sum of the following series a+b,a-b,a-3b... upto 22 terms
Answers
Answered by
96
Answer:
Step-by-step explanation:
n/2 (2a+(n-1)d)
22/2 (2×(a+b)+(22-1)(-2b))
11 (2a+2b+21×(-2b))
11 (2a+2b-42b)
11 (2a-40b)
22a-440b
Answered by
56
a = a+b
n= 22
d=(a-b)-(a+b)=a-b-a-b= --2b
S = n/2{2a+(n -- 1)d}
22/2{2(a+b)+21×--2b
=11(2a+2b-42b)
=11(2a --40b)
=11×2(a -20b)
=22(a --20b)
n= 22
d=(a-b)-(a+b)=a-b-a-b= --2b
S = n/2{2a+(n -- 1)d}
22/2{2(a+b)+21×--2b
=11(2a+2b-42b)
=11(2a --40b)
=11×2(a -20b)
=22(a --20b)
pip2:
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