Sum of n terms of two AP's are in the ratio (3n+5):(5n+3) then ratio of their $10th$ term

Option a: $31:49$. Solution: Let the two A.P.s have first terms a1, a2 and common differences d1, d2 . The ratio of their sum of n terms is: (1)Sn(2) = 2a1 + (n-1)d1/2a2 + (n-1)d2 = 3n+5/5n+3 We want to find the ratio of their $10th terms: 10(1)t10(2) = a1 + 9d1/a2 + 9d2 = 2a1 + 18d1/2a2 + 18d2 By comparing with the sum ratio, we substitute n-1 = 18 n = 19$: 10(1)t10(2) = 3(19) + 5/5(19) + 3 = 57 + 5/95 + 3 = 62/98 = 31/49 Hence, Option A is the correct answer.

Sum of n terms of two AP's are in the ratio (3n+5):(5n+3) then ratio of their $10th$ term

The correct answer is Option a: $31:49$. Let the two A.P.s have first terms a1, a2 and common differences d1, d2 . The ratio of their sum of n terms is: (1)Sn(2) = 2a1 + (n-1)d1/2a2 + (n-1)d2 = 3n+5/5n+3 We want to find the ratio of their $10th terms: 10(1)t10(2) = a1 + 9d1/a2 + 9d2 = 2a1 + 18d1/2a2 + 18d2 By comparing with the sum ratio, we substitute n-1 = 18 n = 19$: 10(1)t10(2) = 3(19) + 5/5(19) + 3 = 57 + 5/95 + 3 = 62/98 = 31/49 Hence, Option A is the correct answer.

Sequence and SeriesMCQMTP June 2023 Series IQuestion 1808 of 213

All Questions

Sum of $\displaystyle n$ terms of two AP's are in the ratio $\displaystyle (3n+5):(5n+3)$ then ratio of their $10^{th}$ term

Select Your Answer (Keyboard: A, B, C, D)

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Correct Answer

✅ Option a — $31:49$

All Options:

A$31:49$
B$30:49$
C$28:49$
DNone of these

Detailed Solution & Explanation

Let the two A.P.s have first terms

\displaystyle a_1, a_2

and common differences

\displaystyle d_1, d_2

.
The ratio of their sum of

\displaystyle n

terms is:

\frac{S_{n}^{(1)}}{S_{n}^{(2)}} = \frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{3 n + 5}{5 n + 3}

We want to find the ratio of their $10^{

\displaystyle \text{th}

}

\displaystyle terms:<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><msubsup><mi>t</mi><mn>10</mn><mrow><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></msubsup><msubsup><mi>t</mi><mn>10</mn><mrow><mo stretchy="false">(</mo><mn>2</mn><mo stretchy="false">)</mo></mrow></msubsup></mfrac><mo>=</mo><mfrac><mrow><msub><mi>a</mi><mn>1</mn></msub><mo>+</mo><mn>9</mn><msub><mi>d</mi><mn>1</mn></msub></mrow><mrow><msub><mi>a</mi><mn>2</mn></msub><mo>+</mo><mn>9</mn><msub><mi>d</mi><mn>2</mn></msub></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>2</mn><msub><mi>a</mi><mn>1</mn></msub><mo>+</mo><mn>18</mn><msub><mi>d</mi><mn>1</mn></msub></mrow><mrow><mn>2</mn><msub><mi>a</mi><mn>2</mn></msub><mo>+</mo><mn>18</mn><msub><mi>d</mi><mn>2</mn></msub></mrow></mfrac></mrow><annotation encoding="application/x-tex">

n-1 = 18 \implies n = 19$:

\frac{t_{10}^{(1)}}{t_{10}^{(2)}} = \frac{3 (19) + 5}{5 (19) + 3} = \frac{57 + 5}{95 + 3} = \frac{62}{98} = \frac{31}{49}

Hence, **Option A** is the correct answer.

About This Chapter: Sequence and Series

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Arithmetic & Geometric Progressions

This chapter covers Arithmetic Progressions (AP) and Geometric Progressions (GP). Students learn how to find the nth term, sum of n terms, arithmetic/geometric means, and sum to infinity of a GP.

View Official ICAI Syllabus

Exam Strategy Tip

For complex 'sum of series' questions, a great hack is to substitute n = 1 and n = 2 into the question and the options to see which one matches, completely bypassing the formula.

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Sum of n\displaystyle nn terms of two AP's are in the ratio (3n+5):(5n+3)\displaystyle (3n+5):(5n+3)(3n+5):(5n+3) then ratio of their $10^{th}$ term