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A gentleman buys every year Banks' certi...

A gentleman buys every year Banks' certificates of value exceeding the last year's purchase by 25. After 20 years he finds that the total value of the certificates purchased by him is 7,250. Find the value of the certificates purchased by him in the 1st year and in the 13th year.

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To solve the problem step by step, we will follow the logic presented in the video transcript. ### Step 1: Define the Variables Let the value of the certificates purchased in the first year be \( a \). Since the value of the certificates increases by 25 every year, the value of the certificates purchased in the second year will be \( a + 25 \), in the third year will be \( a + 50 \), and so on. ### Step 2: Identify the Sequence The values of the certificates purchased over the years form an arithmetic progression (AP): - 1st year: \( a \) - 2nd year: \( a + 25 \) - 3rd year: \( a + 50 \) - ... - 20th year: \( a + 475 \) (since \( 25 \times 19 = 475 \)) ### Step 3: Total Value of Certificates The total value of the certificates purchased over 20 years is given as 7250. The formula for the sum \( S_n \) of the first \( n \) terms of an AP is: \[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \] where: - \( n = 20 \) - \( d = 25 \) Substituting the values into the formula: \[ 7250 = \frac{20}{2} \times (2a + (20 - 1) \times 25) \] \[ 7250 = 10 \times (2a + 475) \] ### Step 4: Simplify the Equation Dividing both sides by 10: \[ 725 = 2a + 475 \] Now, isolate \( 2a \): \[ 2a = 725 - 475 \] \[ 2a = 250 \] Now, divide by 2 to find \( a \): \[ a = \frac{250}{2} = 125 \] ### Step 5: Value of Certificates in the 1st Year Thus, the value of the certificates purchased in the first year is: \[ \text{Value in 1st year} = a = 125 \] ### Step 6: Value of Certificates in the 13th Year To find the value of the certificates purchased in the 13th year, we use the formula for the \( n \)-th term of an AP: \[ a_n = a + (n - 1)d \] Substituting \( n = 13 \): \[ a_{13} = 125 + (13 - 1) \times 25 \] \[ a_{13} = 125 + 12 \times 25 \] \[ a_{13} = 125 + 300 = 425 \] ### Final Answers - The value of the certificates purchased in the 1st year is **125**. - The value of the certificates purchased in the 13th year is **425**.
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ICSE-SEQUENCE AND SERIES -EXERCISE 14 (c)
  1. The interior angles of a polygon are in arithmetic progression. The sm...

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  2. Determine the sum of first 35 terms of an A.P. if t(2), = 1 and t(7) ,...

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  3. Find the sum of all natural numbers between 100 and 1000 which are mul...

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  4. How many terms of the A.P. 1,4,7.... are needed to give the sum 715?

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  5. Find the rth term of an A.P., sum of whose first n terms is 2n + 3n^(2...

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  6. In an arithmetical progression, the sum of p terms is m and the sum of...

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  7. The sum of the first fifteen terms of an arithmetical progression is 1...

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  8. The sum of the first six terms of an arithmetic progression is 42. The...

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  9. A sum of रु6240 is paid off in 30 instalments, such that each instalme...

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  10. The nth term of an A.P. is p and the sum of the first n term is s. Pro...

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  11. The sum of the first n terms of the arithmetical progression 3, 5(1)/(...

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  12. If the sum of the first 4 terms of an arithmetic progression is p, the...

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  13. The last term of an A.P. 2, 5, 8, 11, .... is .x. The sum of the terms...

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  14. A gentleman buys every year Banks' certificates of value exceeding the...

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  15. If the sums of the first n terms of two A.P.'s are in the ratio 7n-5: ...

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  16. If the ratio of the sum of m terms and n terms of an A.P. be m^2 : n^2...

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  17. Let a(1) , a(2) , a(3) , ..... be terms of an A.P. If (a(1)+a(2)+........

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  18. If the sum of n, 2n, 3n terms of an A.P are S(1), S(2), S(3), respecti...

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  19. If the sum of p terms of an A.P. is q and the sum of q terms is p, sho...

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  20. The ratio between the sum of n terms of two A.P.'s is (7n + 1) : (4n+2...

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