CAT 2024 Slot 1 · Quant · Question 14
CAT 2024 Slot 1 previous year question on Sequences and Series, with the answer and a step-by-step solution.
- A204
- B-194
- C-196
- D206
Answer & solution
Answer: Option B
Write the progression as {x}_{k}=a+\left( k-1 \right)d.
The first condition gives a+4d=-4.
Expand the second condition: 2\left( a+5d \right)+2\left( a+8d \right)=\left( a+10d \right)+\left( a+12d \right).
The left side is 4a+26d and the right side is 2a+22d.
So 4a+26d=2a+22d, which reduces to 2a+4d=0, that is a=-2d.
Substituting into a+4d=-4: -2d+4d=-4, so 2d=-4 and d=-2.
Then a=-2\left( -2 \right)=4.
Finally {x}_{100}=a+99d=4+99\left( -2 \right)=4-198=-194.
Shortcut. The second condition is pure structure — it collapses to a=-2d regardless of the first. Simplify it before using {x}_{5}, and the progression is pinned down by one substitution.
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