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Show that in an A.P. the sum of the term...

Show that in an A.P. the sum of the terms equidistant from the beginning and end is always same and equal to the sum of first and last terms.

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Let Ap `be a,a,+d,a+2d *** a+(n-1)d `
`therefore a_(1) +a_(n)=a+a+(n-1)d`
` =2a+(n-1)d`
`Now a_(2) +a_(n-1)=(a+d)+[a+(n-2)d]` ...(i)
` =2a+(n-1)d`
`a_(2)+a_(n-1)=a_(1)+a_(n)` [using Eq ,(i)]
`a_(3)+a_(n-2)=(a+2d)+[a+(n-3)d]`
`=2a+(n-1)d`
`=a_(1)+a_(n)`[using Eq.(i) ]`
Follow this pattern , we see that sum fo terms equidistant from the begining and end in an AP is equal to [first term+term].
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