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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

Valid until 28th Sep

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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

T_{1} = (4 + 3) (4 + 7) = 7 × 11

T_{2} = 11 × 15

T_{3} = 15 × 19 and so on

Expanding T_{n} = (4n + 3) (4n + 7)

We will get 16n^{2} + 12n + 28n +21

16n^{2} + 40n +21

Σ16n^{2} + 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"**

Choice A is the correct answer.

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