Question

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows:rupees 200 for the first day, rupees 250 for the secondary rupees,300 for the third day, etc;at the penalty for each succeeding day being rupees 50 more than for the preceding day. How much does the delay of 30 days cost the contractor ?​

Answer 1

Answer:

A delay of 30 days will cost the contractor of Rs. 27750.

Step-by-step explanation:

Since the penalty for each succeeding day is Rs 50 more than for the preceding day. Therefore, amount of penalty for different days forms an A.P. with first a (=200) and common difference d (=50).

Sn = n/2 [ 2a + (n - 1)d]

Required sum = 30/2 { 2 (200) + (30 - 1)50}

= 15( 400 + 29 × 50)

= 15 (400 +1450)

= 15 × 1850

= 27750

Thus, a delay of 30 days will cost the contractor of Rs. 27750

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Answer 2

$\small\sf\underline\red{The\:amount\:paid\:in\:the\:form\:of\:penalty\:is\:in}$

$\small\sf\underline\red{the\:form\:of\:following\:AP:}$

$\large\sf\underline\red{Rs.200, Rs. 250, Rs.300, Rs. 350}$

$\large\tt\pink{a=200,}$

$\large\tt\pink{d=250-200=50}$

$\large\tt\pink{n=30}$

$\small\sf\underline\red{The\:sum\:of\:n\:terms\:of\:an\:AP\:are\:given\:by}$

$\longrightarrow$$\large\tt\purple{Sn=\frac{n}{2}(2a+(n-1)d)}$

$\longrightarrow$$\large\tt\purple{S30=\frac{30}{2}(2(200)+(30-1)50)}$

$\longrightarrow$$\large\tt\purple{\frac{30}{2}(400+1450)}$

$\longrightarrow$$\large\tt\purple{\frac{30}{2}(1850)}$

$\longrightarrow$$\large\tt\purple{15(1850)}$

$\longrightarrow$$\large{\boxed{\tt{\purple{27750}}}}$