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

a_{1} + a_{n} = a + a + (n – 1)d

a_{1} + a_{n} = 2a + (n – 1)d …(i)

Taking second and second last term

a_{2} + a_{n–2} = (a + d) + [a + (n – 2)d]

= a + d + a + nd – 2d

= 2a + nd – d

= 2a + (n – 1)d

= a_{1} + a_{n} [from (i)]

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

a_{3} + a_{n-3} = (a + 2d) + [a + (n – 3)d]

= a + 2d + a + nd – 3d

= 2a + nd – d

= 2a + (n – 1)d

= a_{1} + a_{n} [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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(ii) are in AP.

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