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

1. 80707
2. 80751
3. 80730
4. 80773

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##### 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
Tn = (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 Tn = (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

The question is "The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + ..... + 95 x 99 is"

##### Hence, the answer is 80707

Choice A is the correct answer.

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