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CAT 2018 Question Paper | Quants Slot 2

CAT Previous Year Paper | CAT Sequence and Series Questions | Question 16

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You can expect at least one question from Sequences and Series topic in the CAT Exam. This CAT 2018 Question Paper is no exception. Make sure you are doubly thorough with your concepts while taking up the CAT Online Preparation. CAT Quants require some specific strateges, and to learn about the preparation techniques, click: How to prepare for Quant.

Question 16 : The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + ..... + 95 x 99 is

80707
80751
80730
80773

Video Solution

Explanatory Answer

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Explanatory Answer

Method of solving this CAT Question from Sequence and Series

Here we have to find the value of the sum 7 × 11 + 11 × 15 + 15 × 19 + ..... + 95 × 99
T_n = (4n + 3) (4n + 7)
Since increment of 4 takes place in every value , 4n is the term to be used
T₁ = (4 + 3) (4 + 7) = 7 × 11
T₂ = 11 × 15
T₃ = 15 × 19 and so on
Expanding T_n = (4n + 3) (4n + 7)
We will get 16n² + 12n + 28n +21
16n² + 40n +21
Σ16n² + 40 Σn + 21 Σ1
\\frac{16n(n+1)(2n+1)}{6}) + \\frac{40n(n+1)}{2}) + 21 × n
Simplifying this we can take n out
n[\\frac{8n(n+1)(2n+1)}{3}) + 20(n+1) + 21]
From the options we can take 80707 and check whether it is divisible by 23
\\frac{80707}{23}) = 3509
⟹ option a) 80707 is a multiple of 23
⟹ option b) 80751 is 80707 + 44 so this doesn’t work
⟹ option c) 80730 is 80707 + 23 so this is also a multiple of 23
⟹ option d) 80773 is 80707 + 66 so this also doesn’t work
So let us check out with option a and c so we have to substitute and simplify and find
n[\\frac{8n(n+1)(2n+1)}{3}) + 20(n+1) + 21]
\\frac{n}{3})[8(n+1)(2n+1) + 60(n+1) + 63]
From the answer choices we have only two possibilities left out i.e. 23 ⨯ 3509 or 23 ⨯ 3510
⟹ \\frac{23}{3})[8(23 + 1)(2(23) + 1) + 60(23 + 1) + 63]
⟹ \\frac{23}{3})[8(24)(47) + 60 (24) + 63]
⟹ \\frac{23}{3})[8(24)(47) + 60 (24) + 63]
⟹ 23[8(8)(47) + 60 (8) + 21]
⟹ 80707
Hence the value of the sum 7 × 11 + 11 × 15 + 15 × 19 + ..... + 95 × 99 is 80707