if the sum of the frist p terms of anA.p is equal to the sum of the frist q terms then show that the sum of its frist p+q terns us zero
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Hello dear,
let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero
Hope its help you dude, :-)
let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero
Hope its help you dude, :-)
kunal224:
thanks bro
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