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Arithmetic Progression Terms Sum

CAT 2003 Slot 1 · Quantitative Ability · Easy · Progressions

This is an easy Quantitative Ability question from the CAT 2003 Slot 1 paper. It tests Progressions. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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The sum of 3rd and 15th elements of an arithmetic progression is equal to the sum of 6th, 11th and 13th elements of the same progression. Then which element of the series should necessarily be equal to zero?

  1. A.

    1st

  2. B.

    9th

  3. C.

    12th

  4. D.

    None of the above

Answer

C

Explanation

Let the 1st term be aa and common difference be dd. T3+T15=(a+2d)+(a+14d)=2a+16dT_3 + T_{15} = (a + 2d) + (a + 14d) = 2a + 16d. T6+T11+T13=(a+5d)+(a+10d)+(a+12d)=3a+27dT_6 + T_{11} + T_{13} = (a + 5d) + (a + 10d) + (a + 12d) = 3a + 27d. Given 2a+16d=3a+27d  ⟹  a+11d=02a + 16d = 3a + 27d \implies a + 11d = 0. T12=a+11d=0T_{12} = a + 11d = 0. Thus, the 12th element must be equal to zero.

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