Sequence and SeriesMCQPYQ Dec 23Question 1776 of 212
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Find the 17th\displaystyle 17^{th} term of an AP series if 15th\displaystyle 15^{th} and 21st\displaystyle 21^{st} terms are 30.5\displaystyle 30.5 and 39.5\displaystyle 39.5 respectively.

Options

A33.5\displaystyle 33.5
B35.5\displaystyle 35.5
C36.0\displaystyle 36.0
D38.0\displaystyle 38.0
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Correct Answer

Option a33.5\displaystyle 33.5

All Options:

  • A33.5\displaystyle 33.5
  • B35.5\displaystyle 35.5
  • C36.0\displaystyle 36.0
  • D38.0\displaystyle 38.0

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Detailed Solution & Explanation

Let the first term be a\displaystyle a and the common difference be d\displaystyle d.
Given:
t15=a+14d=30.5t_{15} = a + 14d = 30.5
t21=a+20d=39.5t_{21} = a + 20d = 39.5
Subtracting the first equation from the second:
6d=9.0    d=1.56d = 9.0 \implies d = 1.5
Substitute d=1.5\displaystyle d = 1.5 back into the first equation:
a+14(1.5)=30.5    a+21=30.5    a=9.5a + 14(1.5) = 30.5 \implies a + 21 = 30.5 \implies a = 9.5

Now, find the 17th\displaystyle 17^{\text{th}} term:
t17=a+16d=9.5+16(1.5)=9.5+24=33.5t_{17} = a + 16d = 9.5 + 16(1.5) = 9.5 + 24 = 33.5
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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