Sequence and SeriesMCQMTP Nov 21Question 1796 of 212
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The sum of the first 3\displaystyle 3 terms in an AP is 18\displaystyle 18 and that of the last 3\displaystyle 3 is 28\displaystyle 28. If the AP has 13\displaystyle 13 terms, find sum of the middle three terms?

Options

A23
B18
C19
Dnone of these
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Correct Answer

✅ Option a — 23

All Options:

  • A23
  • B18
  • C19
  • Dnone of these

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Detailed Solution & Explanation

Let the AP have first term a\displaystyle a and common difference d\displaystyle d. The number of terms is 13\displaystyle 13.
The sum of the first 3\displaystyle 3 terms is 18\displaystyle 18:
t1+t2+t3=a+(a+d)+(a+2d)=3a+3d=18  ⟹  a+d=6— (1)t_1 + t_2 + t_3 = a + (a+d) + (a+2d) = 3a + 3d = 18 \implies a + d = 6 \quad \text{--- (1)}

The sum of the last 3\displaystyle 3 terms (t11,t12,t13\displaystyle t_{11}, t_{12}, t_{13}) is 28\displaystyle 28:
t11+t12+t13=(a+10d)+(a+11d)+(a+12d)=3a+33d=28  ⟹  a+11d=283— (2)t_{11} + t_{12} + t_{13} = (a+10d) + (a+11d) + (a+12d) = 3a + 33d = 28 \implies a + 11d = \frac{28}{3} \quad \text{--- (2)}

Subtracting (1) from (2):
10d=283−6=103  ⟹  d=1310d = \frac{28}{3} - 6 = \frac{10}{3} \implies d = \frac{1}{3}
Substitute d=13\displaystyle d = \frac{1}{3} into (1):
a=6−13=173a = 6 - \frac{1}{3} = \frac{17}{3}

The middle three terms of a 13\displaystyle 13-term A.P. are the 6th\displaystyle 6^{\text{th}}, 7th\displaystyle 7^{\text{th}}, and 8th\displaystyle 8^{\text{th}} terms.
Sum=t6+t7+t8=3t7=3(a+6d)=3(173+6(13))=17+6=23\text{Sum} = t_6 + t_7 + t_8 = 3t_7 = 3(a + 6d) = 3\left(\frac{17}{3} + 6\left(\frac{1}{3}\right)\right) = 17 + 6 = 23
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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