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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