Answer
Verified
402.9k+ views
Hint: Find the smallest and largest 3 digit number when divided by 3 leaves as 2. Then form the arithmetic progression of such numbers. Find first term, last term common difference and number of terms. Substitute these values in the formula to find sum of n terms which is given by \[{{S}_{n}}=\dfrac{n}{2}\left( a+l \right)\], where n is number of terms, a is first term and l is last term.
Complete step by step solution:
The smallest 3 digit number we know is 100. But the smallest 3 digit number that will give us a remainder 2, when divided by 3 is 101.
Similarly the largest 3 digit number is 999. But the largest 3 digit number that will give us a remainder 2, when divided by 3 is 998.
Thus we can say that the lowest 3 digit number is 101 and the highest number is 998. The next number that will leave a remainder of 2 when divided by 3 after 101 is 104 and the next is 107 etc. Thus we can form an AP with it,
101, 104, 107,……. 998
Thus in the Arithmetic progression (AP) we know the first term, a = 101 and the last term, l = 998.
The common difference = \[{{2}^{nd}}\] term - \[{{1}^{st}}\] term = 104 – 101 = 3.
We know the formula for sum of AP as,
= (first term + last term) \[\times \] ${\dfrac{\text{number of terms}}{2}}$.
Sum of n terms, \[{{S}_{n}}=\left( \dfrac{a+l}{2} \right)n\].
We know that the last term of an AP is given as,
\[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\]
We know last term, \[l={{a}_{n}}=998,{{a}_{1}}=101,d=3\]
\[\begin{align}
& \therefore 998=101+\left( n-1 \right)\times 3 \\
& 998-101=3\left( n-1 \right) \\
\end{align}\]
Let us find the value of n.
\[n=299+1=300\]
Thus we got the number of terms in the series, n = 300.
Sum of an AP, \[{{S}_{n}}=\left( \dfrac{a+l}{2} \right)n\]
\[\therefore {{S}_{n}}=\left( \dfrac{101+998}{2} \right)\times 300=\dfrac{1099\times 300}{2}=164850\]
Thus we got the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.
\[\therefore \] Option (d) is the correct answer.
Note: For solving questions like these, the basic concept of Arithmetic progression and its formula should be known. If you know the formulas then they are the direct application of the value and solving it. Since we had already obtained common difference d = 3 and n = 300, we could also have used formula for sum as \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\].
Complete step by step solution:
The smallest 3 digit number we know is 100. But the smallest 3 digit number that will give us a remainder 2, when divided by 3 is 101.
Similarly the largest 3 digit number is 999. But the largest 3 digit number that will give us a remainder 2, when divided by 3 is 998.
Thus we can say that the lowest 3 digit number is 101 and the highest number is 998. The next number that will leave a remainder of 2 when divided by 3 after 101 is 104 and the next is 107 etc. Thus we can form an AP with it,
101, 104, 107,……. 998
Thus in the Arithmetic progression (AP) we know the first term, a = 101 and the last term, l = 998.
The common difference = \[{{2}^{nd}}\] term - \[{{1}^{st}}\] term = 104 – 101 = 3.
We know the formula for sum of AP as,
= (first term + last term) \[\times \] ${\dfrac{\text{number of terms}}{2}}$.
Sum of n terms, \[{{S}_{n}}=\left( \dfrac{a+l}{2} \right)n\].
We know that the last term of an AP is given as,
\[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\]
We know last term, \[l={{a}_{n}}=998,{{a}_{1}}=101,d=3\]
\[\begin{align}
& \therefore 998=101+\left( n-1 \right)\times 3 \\
& 998-101=3\left( n-1 \right) \\
\end{align}\]
Let us find the value of n.
\[n=299+1=300\]
Thus we got the number of terms in the series, n = 300.
Sum of an AP, \[{{S}_{n}}=\left( \dfrac{a+l}{2} \right)n\]
\[\therefore {{S}_{n}}=\left( \dfrac{101+998}{2} \right)\times 300=\dfrac{1099\times 300}{2}=164850\]
Thus we got the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.
\[\therefore \] Option (d) is the correct answer.
Note: For solving questions like these, the basic concept of Arithmetic progression and its formula should be known. If you know the formulas then they are the direct application of the value and solving it. Since we had already obtained common difference d = 3 and n = 300, we could also have used formula for sum as \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\].
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
Write an application to the principal requesting five class 10 english CBSE
What is the type of food and mode of feeding of the class 11 biology CBSE
Name 10 Living and Non living things class 9 biology CBSE