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In an A. P. : given a = 2, d = 8, S (n...

In an A. P. :
given `a = 2, d = 8, S _(n) = 90,` find n and `a _(n)`

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To solve the problem step by step, we will use the formulas related to Arithmetic Progressions (A.P.). ### Given: - First term \( a = 2 \) - Common difference \( d = 8 \) - Sum of the first \( n \) terms \( S_n = 90 \) ### Step 1: Use the formula for the sum of the first \( n \) terms of an A.P. The formula for the sum of the first \( n \) terms \( S_n \) of an A.P. is given by: \[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \] ### Step 2: Substitute the known values into the formula. Substituting \( S_n = 90 \), \( a = 2 \), and \( d = 8 \) into the formula: \[ 90 = \frac{n}{2} \times (2 \times 2 + (n - 1) \times 8) \] This simplifies to: \[ 90 = \frac{n}{2} \times (4 + (n - 1) \times 8) \] ### Step 3: Simplify the equation. Now, simplify the expression inside the parentheses: \[ 90 = \frac{n}{2} \times (4 + 8n - 8) \] \[ 90 = \frac{n}{2} \times (8n - 4) \] \[ 90 = \frac{n(8n - 4)}{2} \] Multiply both sides by 2 to eliminate the fraction: \[ 180 = n(8n - 4) \] \[ 180 = 8n^2 - 4n \] ### Step 4: Rearrange the equation into standard quadratic form. Rearranging gives: \[ 8n^2 - 4n - 180 = 0 \] ### Step 5: Simplify the quadratic equation. We can simplify the equation by dividing all terms by 4: \[ 2n^2 - n - 45 = 0 \] ### Step 6: Factor the quadratic equation. To factor \( 2n^2 - n - 45 \), we look for two numbers that multiply to \( 2 \times -45 = -90 \) and add to \( -1 \). The numbers are \( -10 \) and \( 9 \): \[ 2n^2 - 10n + 9n - 45 = 0 \] Grouping gives: \[ 2n(n - 5) + 9(n - 5) = 0 \] Factoring out \( (n - 5) \): \[ (n - 5)(2n + 9) = 0 \] ### Step 7: Solve for \( n \). Setting each factor to zero gives: 1. \( n - 5 = 0 \) → \( n = 5 \) 2. \( 2n + 9 = 0 \) → \( n = -\frac{9}{2} \) (not valid since \( n \) must be positive) Thus, \( n = 5 \). ### Step 8: Find the \( n \)-th term \( a_n \). The formula for the \( n \)-th term \( a_n \) is: \[ a_n = a + (n - 1)d \] Substituting \( n = 5 \): \[ a_5 = 2 + (5 - 1) \times 8 \] \[ a_5 = 2 + 4 \times 8 \] \[ a_5 = 2 + 32 = 34 \] ### Final Answers: - \( n = 5 \) - \( a_n = 34 \)
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OSWAL PUBLICATION-ARITHMETIC PROGRESSIONS -NCERT CORNER (EXERCISE-5.3)
  1. given: a3=15,S[10]=125, find d and a[10]

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  2. given d = 5, S9 = 75, find a and a9.

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  3. In an A. P. : given a = 2, d = 8, S (n) = 90, find n and a (n)

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  4. In an A. P. : given a = 8, a (n) = 62, S (n) = 210, find n and d.

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  5. In an AP, given an=4, d=2, Sn=-14 find n and a

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  6. In an A. P. : given a = 3, n = 8, S = 192, find d.

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  7. In an A. P. : given l = 28, S = 144, and there are total 9 terms Fin...

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  8. How many terms of the AP: 9, 17, 25, . . . must be taken to give a sum...

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  9. The first term of an A.P. is 5, the last term is 45 and the sum is 400...

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  10. The first and the last terms of an A.P. are 17 and 350 respectively. I...

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  11. Find the sum of first 22 terms of an A.P. in which d = 7 and 22 nd ter...

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  12. Find the sum of first 51 terms of an A.P. whose second and third terms...

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  13. If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 2...

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  14. Show that a (1), a (2), …., a (n),… from an A.P. where a (n) is define...

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  15. Show that a (1), a (2), …., a (n),… from an A.P. where a (n) is define...

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  16. If the sum of the first n terms of an A.P. is 4n - n ^(2), what is the...

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  17. The sum of first 40 positive integers divisible by 6 is

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  18. Find the sum of the first 15 multiples of 8.

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  19. Find the sum of the odd numbers between 0 and 50.

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  20. A contract on construction job specifies a penalty for delay of com...

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