Question

plz help me to solve this question ​

Answer 1

Answer:

In 16 rows, 200 logs are placed and the top row has 5.

Step-by-step explanation:

Let the First row = a1 = 20 logs

So, Common difference = a2 - a1 = 19 - 20 = -1

Finding the nth term-

an = a1 + (n - 1)d

an = 20 + (n - 1)(-1)

an = 20 - n + 1

an = 21 - n

Now, Finding nth term:

Sn = 200

Sn = n/2 (a1 + an)

200 = n/2 (20 + 21 - n)

400 = n(41 - n)

400 = 41n - n²

n² - 41n + 400 = 0

(n - 16)(n - 25) = 0

n = 16 or n = 25

Now,

Assuming n = 16

a16 = 21 - 16

a16 = 5

Assuming n = 25

an = 21 - 25 = -4

an cannot be negative, so an = -4 can't be accepted.

Answer: an = 5

Final Answer: In 16 rows, 200 logs are placed and the top row has 5.

Answer 2

Question :-- 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row,18 in the row next to it and so on . In how may rows are the 200 logs placedand how many logs are in the top row ?

Concept used :---

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

______________________________

solution :---

As, the rows are going up, the no of logs are decreasing,

20, 19, 18, .......

it's an AP series .

Suppose 200 logs are arranged in 'n' rows,

then We have:----

→ First term, a = 20,

→ Common difference, d = - 1

→ Sum of n terms, Sn = No of logs = 200

Putting values in sum formula we get,

→ 200 = n/2[2*20+(n-1)(-1)]

→ 400 = n [ 40 - n +1 ]

→ 400 = 41n - n²

→ n² - 41n + 400 = 0

Solving the Equation by splitting the middle term now,

→ n² - 16n - 25n + 400 = 0

→ n(n-16) -25(n-16) = 0

→ (n-16)(n-25) = 0

Putting both Equal to zero we get,

→ n -16 = 0. or, n-25 = 0

→ n = 16 or, n = 25

___________________________

Now, when Number of Rows are 25, than

→ an = 20+(25-1)(-1)

→ an = 20 -24

→ an = (-4)

Since negative value for number of logs is not possible hence, number of rows = 16 ..

→ an = 20+(16-1)(-1)

→ an = 20 -15

→ an = 5 logs.

Hence, Number of rows are 16 and Number of logs in Top row are 5 .