Q. 43.6( 145 Votes )

# The houses of a row are numbered consecutively from 1 to 49. Show that there is a valueof x such that the sum of the numbers of the houses preceding the house numbered x isequal to the sum of the numbers of the houses following it. Find this value of x.

[Hint :]

Answer :

The AP in the above problem is

1, 2, 3, - - - , 49

With first term, a = 1

Common difference, d = 1

nth term of AP = a + (n - 1)d

a_{n} = 1 + (n - 1)1

a_{n} = n [1]

Suppose there exist a mth term such that, (m<49)

Sum of first m - 1 terms of AP = Sum of terms following the mth term

Sum of first m - 1 terms of AP = Sum of whole AP - Sum of first m terms of AP

As we know sum of first n terms of an AP is, if last term a_{n} is given

(m - 1)(1 + m - 1) = 49(1 + 49) - m(1 + m) [using 1]

(m - 1)m = 2450 - m(1 + m)

m^{2} - m = 2450 - m + m^{2}

2m^{2} = 2450

m^{2} = 1225

m = 35 or m = - 35 [not possible as no of terms can't be negative.]

and a_{m} = m = 35 [using 1]

So, sum of no of houses preceding the house no 35 is equal to the sum of no of houses following the house no 35.

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The sum of n terms of an A.P. is 3n^{2}+ 5n. Find the A.P. Hence, find its 16th term.