Sequence and SeriesMCQPYQ Dec 23Question 1778 of 212
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Find the value of 'x\displaystyle x' for the following data 1+7+13+19.........+x=225\displaystyle 1+7+13+19.........+x=225

Options

A56\displaystyle 56
B63\displaystyle 63
C49\displaystyle 49
D42\displaystyle 42
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Correct Answer

✅ Option c — 49\displaystyle 49

All Options:

  • A56\displaystyle 56
  • B63\displaystyle 63
  • C49\displaystyle 49
  • D42\displaystyle 42

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Detailed Solution & Explanation

The given series is 1+7+13+19+⋯+x=225\displaystyle 1 + 7 + 13 + 19 + \dots + x = 225.
This is an A.P. with:
First term a=1\displaystyle a = 1
Common difference d=6\displaystyle d = 6
Sum Sn=225\displaystyle S_n = 225.

Using the sum formula:
Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} [2a + (n-1)d]
225=n2[2(1)+(n−1)6]225 = \frac{n}{2} [2(1) + (n-1)6]
225=n[1+3n−3]225 = n [1 + 3n - 3]
225=n(3n−2)225 = n(3n - 2)
3n2−2n−225=03n^2 - 2n - 225 = 0
Solving this quadratic equation:
3n2−27n+25n−225=03n^2 - 27n + 25n - 225 = 0
3n(n−9)+25(n−9)=03n(n - 9) + 25(n - 9) = 0
(3n+25)(n−9)=0(3n + 25)(n - 9) = 0
Since n\displaystyle n must be a positive integer, n=9\displaystyle n = 9.

Now, find the value of x\displaystyle x (the 9th\displaystyle 9^{\text{th}} term):
x=t9=a+8d=1+8(6)=1+48=49x = t_9 = a + 8d = 1 + 8(6) = 1 + 48 = 49
Hence, **Option C** 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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