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Arithmetic Progression Term

CAT 2024 Slot 1 · Quantitative Ability · Medium · Progressions

This is a medium Quantitative Ability question from the CAT 2024 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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Suppose x1,x2,x3,…,x100x_1, x_2, x_3, \dots, x_{100} are in arithmetic progression such that x5=−4x_5 = -4 and 2x6+2x9=x11+x132x_6 + 2x_9 = x_{11} + x_{13}. Then, x100x_{100} equals

  1. A.

    204

  2. B.

    -194

  3. C.

    -196

  4. D.

    206

Answer

B

Explanation

Let first term be aa and common difference be dd. x5=a+4d=−4x_5 = a + 4d = -4.

Given 2x6+2x9=x11+x132x_6 + 2x_9 = x_{11} + x_{13}: 2(a+5d)+2(a+8d)=(a+10d)+(a+12d)2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d) 4a+26d=2a+22d  ⟹  2a+4d=0  ⟹  a=−2d4a + 26d = 2a + 22d \implies 2a + 4d = 0 \implies a = -2d

Substitute a=−2da = -2d into a+4d=−4a + 4d = -4: −2d+4d=−4  ⟹  2d=−4  ⟹  d=−2-2d + 4d = -4 \implies 2d = -4 \implies d = -2 a=4a = 4

Then x100=a+99d=4+99(−2)=−194x_{100} = a + 99d = 4 + 99(-2) = -194.

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