We can use the formula to find the sum of 'n' terms of an arithmetic series. The above sequence is an arithmetic progression. If we write the number of bricks in each row as a sequence, we get The construction has to be stopped when it reaches the 25 th row. The longer side of trapezoid shaped garden is containing 97 and each row must be decreased by 2. Each row must be decreased by 2 bricks on each end and the construction should stop at 25 th row. The longer side of trapezoid needs to start with a row of 97 bricks. Therefore the clock will strike 156 times in a day.Ī gardener plans to construct a trapezoidal shaped structure in his garden. The above sequences are arithmetic sequences.įind the sum of terms in both the sequences.īecause the sequences are arithmetic progressions, we can use the formula to find sum of 'n' terms of an arithmetic series. The clock strikes once at 1'o clock, twice at 2'o clock and so on.ġ, 2, 3. If a clock strikes once at 1'o clock,twice at 2'o clock and so on. So, the total interest earned at the and of 30 years is $37200. To find the total interest for 30 years, we have to find the sum of 30 terms in the above arithmetic progression.įormula to find sum of 'n' terms in an arithmetic progression is Therefore interest amounts form an arithmetic progression. This sequence is an arithmetic progression. So, the interest amounts from the first year are In the second year, amount deposited is $1000. Then, interest earned at the end of the first year is In the first year, amount deposited is $1000. If so,find the total interest at the end of 30 years.įirst let us find the interest using simple interest formula. Calculate the interest at the end of each year. The sum of $1000 is deposited every year at 8% simple interest. Here 'n' represents number of days delayed. So, we have to write this amount as seriesĪnd the sum of the penalty amount is 165000. Total cost: 80 times the average cost per meter: 80109 8720. The company can pay 165000 as penalty for this delay at maximum. It is the sum of the first 80 terms of an arithmetic progression with the first term of 30 and the common difference of 2. Let us write the penalty amount paid by the construction company from the first day as sequence Find the maximum number of days by which the completion of work can be delayed Based on its budget, the company can afford to pay a maximum of $ 165000 toward penalty. The penalty will be $4000 for the first day and will increase by $10000 for each following day. A construction company will be penalized each day of delay in construction for bridge.
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