CAT 2021 Slot 2 · Quant · Question 20
CAT 2021 Slot 2 previous year question on Sequences and Series, with the answer and a step-by-step solution.
- A2
- B-2
- C200
- D-200
Answer & solution
Answer: Option B
Call the given alternating sum {S}_{n}, so {S}_{n}={n}^{2}+2n. Consecutive values of S differ by a single term, with the sign that term carries.
{S}_{48}={48}^{2}+96=2400,
{S}_{49}={49}^{2}+98=2499,
{S}_{50}={50}^{2}+100=2600.
In {S}_{49} the last term {x}_{49} carries the sign {\left( -1 \right)}^{50}=+1,
so {S}_{49}-{S}_{48}={x}_{49},
giving {x}_{49}=99.
In {S}_{50} the last term {x}_{50} carries the sign {\left( -1 \right)}^{51}=-1,
so {S}_{50}-{S}_{49}=-{x}_{50}, giving {x}_{50}=-101.
Hence {x}_{49}+{x}_{50}=99-101=-2.
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