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