Question

If \(\frac{5+9+13+…… \text{ to } n \text{ terms}}{7+9+11+……. \text{ to } (n+1) \text{ terms}}\) \(= \frac{17}{16}\) then\(n =\)

• A. 8
• B. 7
• C. 10
• D. 11

Hint:

1. Sum of Numerator AP (\(a=5, d=4\)): \(S_N = \frac{n}{2}(10 + (n-1)4) = n(2n+3)\). 2. Sum of Denominator AP (\(a=7, d=2\), terms=\(n+1\)): \(S_D = \frac{n+1}{2}(14 + n \cdot 2) = (n+1)(n+7)\). 3. Equation: \( \frac{n(2n+3)}{(n+1)(n+7)} = \frac{17}{16} \). 4. Test options. For \(n=7\): Numerator sum = \(7(14+3) = 7(17) = 119\). Denominator sum = \((7+1)(7+7) = 8(14) = 112\). Ratio = \(119/112\). Divide by 7: \( (119/7) / (112/7) = 17/16 \). Matches. So, \(n=7\) is correct.

Answer 1

The Correct Option is B

Concept: This problem involves finding the sum of terms in two different arithmetic progressions (APs) and then solving an equation based on the ratio of these sums. The sum of an AP is given by \( S_k = \frac{k}{2}[2a + (k-1)d] \), where \(k\) is the number of terms, \(a\) is the first term, and \(d\) is the common difference. Step 1: Find the sum of the numerator AP (\(S_N\)) The numerator series is \(5+9+13+\ldots\) to \(n\) terms. This is an AP with:
First term, \(a_N = 5\)
Common difference, \(d_N = 9-5 = 4\)
Number of terms = \(n\) Sum of numerator, \(S_N = \frac{n}{2}[2a_N + (n-1)d_N]\) \(S_N = \frac{n}{2}[2(5) + (n-1)4]\) \(S_N = \frac{n}{2}[10 + 4n - 4]\) \(S_N = \frac{n}{2}[4n + 6]\) \(S_N = n(2n + 3)\) Step 2: Find the sum of the denominator AP (\(S_D\)) The denominator series is \(7+9+11+\ldots\) to \((n+1)\) terms. This is an AP with:
First term, \(a_D = 7\)
Common difference, \(d_D = 9-7 = 2\)
Number of terms = \((n+1)\) Sum of denominator, \(S_D = \frac{n+1}{2}[2a_D + ((n+1)-1)d_D]\) \(S_D = \frac{n+1}{2}[2(7) + (n)(2)]\) \(S_D = \frac{n+1}{2}[14 + 2n]\) \(S_D = \frac{n+1}{2} \cdot 2(7 + n)\) \(S_D = (n+1)(n+7)\) Step 3: Set up the equation using the given ratio We are given that \( \frac{S_N}{S_D} = \frac{17}{16} \). Substitute the expressions for \(S_N\) and \(S_D\): \[ \frac{n(2n+3)}{(n+1)(n+7)} = \frac{17}{16} \] Step 4: Solve for \(n\) Cross-multiply: \[ 16n(2n+3) = 17(n+1)(n+7) \] Expand both sides: \[ 16(2n^2 + 3n) = 17(n^2 + 7n + n + 7) \] \[ 32n^2 + 48n = 17(n^2 + 8n + 7) \] \[ 32n^2 + 48n = 17n^2 + 136n + 119 \] Bring all terms to one side to form a quadratic equation: \[ 32n^2 - 17n^2 + 48n - 136n - 119 = 0 \] \[ 15n^2 - 88n - 119 = 0 \] We can solve this quadratic equation or test the given options. Let's test \(n=7\) (option 2, which is circled in the image). If \(n=7\): Left side: \(15(7)^2 - 88(7) - 119\) \( = 15(49) - 616 - 119 \) \( = 735 - 616 - 119 \) \( = 735 - (616 + 119) \) \( = 735 - 735 = 0 \) Since the equation holds true for \(n=7\), this is a valid solution. As \(n\) represents the number of terms, it must be a positive integer. The other root of this quadratic equation is \(n = -119/(15 \times 7) = -17/15\), which is not a natural number. Thus, \(n=7\) is the correct answer. Alternatively, after Step 3, with the equation \( \frac{n(2n+3)}{(n+1)(n+7)} = \frac{17}{16} \), we could directly substitute \(n=7\): Numerator for \(n=7\): \(S_N = 7(2(7)+3) = 7(14+3) = 7(17) = 119\). Denominator for \(n=7\): \(S_D = (7+1)(7+7) = (8)(14) = 112\). Ratio: \( \frac{119}{112} \). To simplify \(\frac{119}{112}\), we can see if they have common factors. Both are divisible by 7: \(119 \div 7 = 17\) \(112 \div 7 = 16\) So, \(\frac{119}{112} = \frac{17}{16}\). This matches the given ratio.