The first and last terms of an AP are a and l repectively. Show that the sum of hte nth term from the beginning and the nth tern from the end is (a + 1).

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Solution

Given, the first term, = a

Last term, = l

nth term from the beginning, $a_{n} = a + (n - 1) d$

nth term from the end, $t_{n} = (I - (n - 1) d)$

Now, = [ a + (n – 1) d ] + [ l – (n – 1) d ]

= (a + l)

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The first and last terms of an AP are a and l repectively. Show that the sum of hte nth term from the beginning and the nth tern from the end is (a + 1).

Given, the first term, = a Last term, = l nth term from the beginning, an=a+(n−1)d nth term from the end,tn=(I−(n−1)d) Now, = [ a + (n – 1) d ] + [ l – (n – 1) d ] = (a + l)

Given, the first term, = a

Last term, = l

nth term from the beginning, $a_{n} = a + (n - 1) d$

nth term from the end, $t_{n} = (I - (n - 1) d)$

Now, = [ a + (n – 1) d ] + [ l – (n – 1) d ]

= (a + l)