The first and the last terms of an A.P. are a and l respectively. Show that the sum of nth term from the beginning and nth term from the end is a + l.

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Solution

Given:
First term = a
Last term = l
nth term from the beginning = $a + (n - 1) d$ , where d is the common difference.
nth term from the end = $l + (n - 1) (- d) = l - d n + d$
Their sum = $a + (n - 1) d + l - d n + d$
$= a + n d - d + l - n d + d = a + l$
Hence, proved.

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The first and the last terms of an A.P. are a and l respectively. Show that the sum of nth term from the beginning and nth term from the end is a + l.

Given: First term = a Last term = l nth term from the beginning = a+(n-1)d, where d is the common difference. nth term from the end = l+(n-1)(-d) =l-dn+d Their sum = a+(n-1)d +l-dn+d = a+nd-d+l-nd+d = a+l Hence, proved.

Given:
First term = a
Last term = l
nth term from the beginning = $a + (n - 1) d$ , where d is the common difference.
nth term from the end = $l + (n - 1) (- d) = l - d n + d$
Their sum = $a + (n - 1) d + l - d n + d$
$= a + n d - d + l - n d + d = a + l$
Hence, proved.