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प्रश्न
The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of x such that sum of numbers of houses preceding the house numbered x is equal to sum of the numbers of houses following x.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint: `S_(x - 1) = S_49 - S_x`]
उत्तर १
Let there be a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it.
That is, 1 + 2 + 3 ... + (x - 1) = (x + 1) + (x + 2) + ... + 49
∴ 1 + 2 + 3 + ... + (x - 1)
= [1 + 2 + ... + x + (x + 1) + ... + 49] - (1 + 2 + 3 + ... + x)
∴ `x - 1/2 [1 + x - 1] = 49/2 [1 + 49] - x/2 [1 + x]`
∴ x(x - 1) = 49 × 50 - x(1 + x)
∴ x(x - 1) + x(1 + x) = 49 × 50
∴ x2 - x + x + x2
= 49 × 50
∴ x2 = 49 × 25
∴ x = 7 × 5
∴ x = 35
Since x is not a fraction, the value of x satisfying the given condition exists and is equal to 35.
उत्तर २
Let there be a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it.
a = 1 and d = 2 - 1 = 1 and n = 49
∴ `S_(x-1)` = S49 - Sx
⇒ `(x - 1)/2[2a + (x - 2)d]`
= `49/2[2a + (49 - 1)d] - x/2[2a + (x - 1) xx d]`
⇒ `(x - 1)/2[2 xx 1 + (x - 2) xx 1]`
= `49/2[2 xx 1 + 48 xx 1] - x/2[2 xx 1 + (x - 1) xx 1]`
⇒ (x - 1)(x)
= `49(50) - x(x + 1)`
⇒ x2 - x
= 2450 - x2 - x
⇒ 2x2 = 2450
⇒ x2 = `2450/2`
⇒ x2 = 1225
⇒ x = ±`sqrt1225`
⇒ x = ± 35
Since x is not a fraction, the value of x satisfying the given condition exists and is equal to 35.
संबंधित प्रश्न
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Solution: First term = a = 6, common difference = d = 3, S27 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]` - Formula
Sn = `27/2 [12 + (27 - 1)square]`
= `27/2 xx square`
= 27 × 45
S27 = `square`
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