If an A.P. consists of n terms with first term a and nth term l show that the sum of the mth term from the beginning and the mth term from the end is (a+l).

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Solution

Given :
First term of A.P is a
nth term =l
Let d be the common difference
nth term=a+(n-1)d
l=a+(n-1)d
d=(l-a)/(n-1) ------------1

Mth term from the beginning
=a+(m-1)d=a+(m-1)(l-a)/(n-1 ) ----------------2

Mth term from the end we get:
where we consider first term as l and the common difference=-d
=l+(m-1)(-d)
=l+(m-1)x(-(l-a)/n-1)
=I-(m-1)(l-a)/(n-1) --------------3

thus sum of mth term from beginning to end is
=a+(m-1)(l-a/n-1 ) +I-(m-1)(l-a)/n-1
=a+l
Hence proved

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