Question

A boy arrange rows of marbles one againts the other sobthat rach row contains the marble less than the preceeding. The lastbrow consist of one marble only,which forms the apex o an triangle.If the boy has 153 marbles .how many marbles are there in the base o the biggest triangle he can construct

Answer 1

Step-by-step explanation:

n^2+n-306=0

n^2+18n-17n-306=0

n(n+18)-17(n+18)=0

(n-17) (n+18)=0

n=17

Answer 2

The number of marbles at the base of the biggest triangle will be 17.

Given:

Total number of marbles = 153

To Find:

The number of marbles at the base of the biggest triangle.

Solution:

It's fairly simple to find the answer to this question, as seen below.

From this arrangement, it is clear that,

It resembles an arithmetic progression

The first term (a) = 1

The common difference (d) = 1

The total number of marbles (Sn) = 153

We know that,

Sum of an arithmetic sequence,

$S_{n} = \frac{n}{2}$ × $(2a + (n-1) d)$

On substituting the given values,

153 = $\frac{n}{2}$ × $(2 + (n-1) 1)$

153 = $\frac{n}{2}$ × (2+ n - 1)

153 × 2 = n (2+ n - 1)

306 = n (n + 1)

306 = n² + n

That is,

n² + n – 306 = 0

For solving this we have an equation,

$x = \frac{-b +/- \sqrt{b^{2} - 4ac } }{2a}$

We have,

x = n

a = 1

b = 1

c = -306

On substituting the values,

$n = \frac{-1 +/- \sqrt{1^{2} - 4*1*-206 } }{2*1}$

$n = \frac{-1 +/- \sqrt{1 + 1224 } }{2}$

$n = \frac{-1 +/- \sqrt{1225} }{2}$

$n = \frac{-1 +/- 35 }{2}$

$n = \frac{-1 + 35}{2}$ or $n = \frac{-1 -35}{2}$

$n = \frac{34}{2}$ or $n= \frac{-36}{2}$

$n = 17$ or $n = -18$

It cannot be negative.

Hence, the number of marbles at the base of the biggest triangle will be 17.

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