Q. 285.0( 1 Vote )

Fill in the blanks

The sum of terms equidistant from the beginning and end in an A.P. is equal to ............ .

Answer :

Let A.P be a, a + d, a + 2d, …, a + (n – 1)d

Taking first and last term


a1 + an = a + a + (n – 1)d


a1 + an = 2a + (n – 1)d …(i)


Taking second and second last term


a2 + an–2 = (a + d) + [a + (n – 2)d]


= a + d + a + nd – 2d


= 2a + nd – d


= 2a + (n – 1)d


= a1 + an [from (i)]


Taking third term from the beginning and the third from the end


a3 + an-3 = (a + 2d) + [a + (n – 3)d]


= a + 2d + a + nd – 3d


= 2a + nd – d


= 2a + (n – 1)d


= a1 + an [from (i)]


From the above pattern, we see that the sum of terms equidistant from the beginning and end in an A.P. is equal to first term + last term.


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