Question

★ Arithmetic Progressions ★

1: A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows :
• Rs 200 for the first day
• Rs 250 for the second day
• Rs 300 for the third day etc...
the penalty for such succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days ?

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Answer 1

Answer:

₹1500 penalty

Step-by-step explanation:

as the penalty for a day is ₹ 50 ,

so,

for 30 days it would be = 30 × 50

=₹ 1500

hope this helps you❣️

Answer 2

$\huge\bf\pink{★ANSWER:}$

The AP is : 200, 250, 300,...

where

a¹ = 200

d = 50

(a²- a¹= 250 - 200 = 50)

The penalty is :

Sⁿ = n => 30

S³⁰ = Rs. ?

So,

$Sⁿ = \frac{n}{2} [2a+(n-1)d] \\ \\ S³⁰= \frac{30}{2} \: [2(200) + (30 - 1)50] \\ \\ S³⁰ = 15 \: (400 + 29 \times 50) \\ \\ S³⁰ = 15 \: (400 + 1450) \\ \\ S³⁰ = 15 \times 1850 \\ \\ {\boxed{\purple{S³⁰ = 27750}}}$

HOPE IT HELPS

MARK BRAINLIEST PLS :-)