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Virat
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Asked: 2026-09-09T07:26:05+00:00 In:

Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.

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Given a + 10d = 38 and a + 15d = 73, subtracting gives 5d = 35, so d = 7 and a = -32. Using tn = a + (n – 1)d, the 31st term is -32 + 30 x 7 = 178.

31st term of an ap whose 11th term is 38 and 16th term is 73cbse half yearly exam question paper downloadcbse internal assessmentcbse noteshalf yearly exam question paper – class 9 math exploring sequences and progressionsimportant questions predicting what comes next: exploring sequences and progressions
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    1. Let a be the first term and d be the common difference.

      Given:

      11th term: t11 = a + 10d = 38

      16th term: t16 = a + 15d = 73

      Subtracting the first equation from the second:

      (a + 15d) – (a + 10d) = 73 – 38

      5d = 35

      d = 7.

      Substituting d = 7 into the first equation:

      a + 10 x 7 = 38

      a + 70 = 38

      a = 38 – 70 = -32.

      Now, finding the 31st term:

      t31 = a + 30d = -32 + 30 x 7 = -32 + 210 = 178.

      Hence, the 31st term is 178.

       

      For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)

      https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/

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