Math, asked by naziyashaikh, 10 months ago

840 logs of wood are placed in 7 rows
one above the others The bottom row
consists of maximum dogs. Each of the
subsequent rows consists of 30 logs less
than the previous row. Find the number of logs in each row.​

Answers

Answered by adityababan12345
10

Answer:

210,180,150,120,90,60,30.

Step-by-step explanation:

Let the number of logs in first row be = x

No. of rows = 7

Difference in logs of each row = (-30)

This becomes an arithmetic progression,

where a = x ; d = (-30) ; n = 7 ; Sₙ = 840

Now by using the formula for finding the sum of 'n' terms of an A.P., we get,

Sₙ = n/2 x {2a + (n-1)d}

840 = 7/2 x {2x + 6(-30)}

840(2)/7 = 2x - 180

(240 + 180)/2 = x

210 = x

Hence the number of loge in each row were 210,180,150,120,90,60,30.

Hope it helps you....

Answered by Anonymous
72

AnswEr :

Let the Number of Logs in Top Row be 30. & as there is Maximum Logs in Bottom Row, then we will Add the Common Difference i.e. 30 here.

\bigstar\:\:\boxed{\sf\:S_n = \dfrac{n}{2} \times \bigg\lbrack2a+(n-1)d\bigg\rbrack}\\\\ \qquad \qquad \qquad \qquad\sf where\\\bullet\:\:\textsf{$\sf S_n$ = Sum of n terms}\\\bullet\:\:\textsf{n = Number of Terms}\\\bullet\:\:\textsf{a = First Term}\\\bullet\:\:\textsf{d = Common Difference}

\rule{150}{2}

\underline{\bigstar\:\textsf{According to the Question Now :}}

:\implies\tt S_n = \dfrac{n}{2} \times \bigg\lbrack2a+(n-1)d\bigg\rbrack\\\\\\:\implies\tt 840 = \dfrac{n}{2} \times \bigg\lbrack2 \times 30+(n-1) \times 30\bigg\rbrack\\\\\\:\implies\tt 840 = \dfrac{n}{2}\times 2\bigg\lbrack30+(n-1) \times 15\bigg\rbrack\\\\\\:\implies\tt 840 =n\times \bigg\lbrack30+15n-15\bigg\rbrack\\\\\\:\implies\tt 840 =n\times \bigg\lbrack15n + 15\bigg\rbrack\\\\\\:\implies\tt 840 = n \times 15 \times (n + 1)\\\\\\:\implies\tt 56 = n(n + 1)\\\\\\:\implies\tt 56 = n^{2} + n\\\\\\:\implies\tt n^{2} + n - 56 = 0\\\\\\:\implies\tt {n}^{2} + (8 - 7)n - 56 = 0\\\\\\:\implies\tt n^{2} + 8n - 7n - 56 = 0\\\\\\:\implies\tt n(n + 8) - 7(n + 8) = 0\\\\\\:\implies\tt (n - 7)(n + 8) = 0\\\\\\:\implies\tt \green{n = 7} \quad or \quad \red{n = - \:8}

\rule{200}{1}

\underline{\bigstar\:\textsf{Number of logs in each row :}}

\begin{tabular}{|c|c|c|c|c|c|c|c|c}\cline{1-8} Row & 1st & 2nd & 3rd & 4th & 5th & 6th & 7th\\\cline{1-8}Logs&30&60&90&120&150&180&210\\\cline{1-8}\end{tabular}

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