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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Sequences and Series

NDA 2018 · Mathematics · Sequences and Series

What is the sum of all two-digit numbers which when divided by 3...

What is the sum of all two-digit numbers which when divided by 3 leave 2 as the remainder?

Options

  1. A1565
  2. B1585
  3. C1635
  4. D1655

Correct answer

C. 1635

Step-by-step solution

The numbers which divided by 3 , leaving remainder 2 will be of the form 3x+23 x+2 Given, 3x+23 \mathrm{x}+2 is 2 -digit number So, xx can be from 3 to 32  Sum of numbers =∑x=332(3x+2)=3(3+4+5+….+32)+(2+2+….)....(1)3,4,5….32 is an A.P with a=3,d=1,Tn=32∴Tn=a+(n−1)1⇒32=3+(n−1)1⇒n−1=29⇒n=30 terms ∴(1)⇒3(3+4+5+….+32)+(2+2+….30 times )=3(302(3+32))+(2×30)(∵Sn=n2(a+ℓ))=902(35)+60=(45×35)+60=1575+60=1635\begin{aligned} & \text{ Sum of numbers }=\sum_{\mathrm{x}=3}^{32}(3 \mathrm{x}+2) \\ & =3(3+4+5+\ldots .+32)+(2+2+\ldots .)....(1)\\ & 3,4,5 \ldots .32 \text{ is an A.P with } \mathrm{a}=3, \mathrm{ d}=1, \mathrm{ T}_{\mathrm{n}}=32 \\ & \therefore \mathrm{ T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) 1 \\ & \Rightarrow 32=3+(\mathrm{n}-1) 1 \\ & \Rightarrow \mathrm{n}-1=29 \Rightarrow \mathrm{n}=30 \text{ terms } \\ & \therefore(1) \Rightarrow 3(3+4+5+\ldots .+32)+(2+2+\ldots .30 \text{ times }) \\ & =3\left(\frac{30}{2}(3+32)\right)+(2 \times 30)\left(\because \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\ell)\right) \\ & =\frac{90}{2}(35)+60=(45 \times 35)+60 \\ & =1575+60=1635 \end{aligned}
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