
The sum of all two-digit numbers which leave a remainder $1$ when divided by $3$ is?
Answer
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Hint: When solving such a question, we need to identify the pattern between the numbers and decide if they are in an arithmetic progression or a geometric progression or an arithmetic-geometric progression. This will make the calculation easier.
For an AP, the sum of n terms = $\dfrac{n}{2}(2a + (n - 1)d)$
For a GP, the sum of n terms = $\dfrac{{a({r^n} - 1)}}{{r - 1}}$
For an AGP, the sum of n terms = $\dfrac{{a - [a + (n - 1)d]{r^n}}}{{1 - r}} + \dfrac{{dr(1 - {r^{n - 1}})}}{{{{(1 - r)}^2}}}$
Where
$r$= common ratio
$d$= common difference
$a$= first term in the sequence
$n$= number of terms
Complete step by step answer:
The question asks us to find the sum of all two-digit numbers which leave a remainder 1 when divided by 3, first let’s list out all the numbers that follow the above criteria.
The numbers are: 10, 13, 16, … , 97
The above numbers are in an AP with a common difference of 3. The first term is 10 and the last term is 97.
Now for finding the sum we can use the formula for finding sum of n terms in an AP. a = 10, d = 3
For finding the value of n, we can use the general formula of nth term in an AP.
Which is $a + (n - 1)d$
Now after substituting all the values in the formula, we get:
$10 + (n - 1)3 = 97$
$3(n - 1) = 97 - 10 = 87$$(n - 1) = \dfrac{{87}}{3} = 29$
$n = 29 + 1 = 30$
Now we have all the required values to find the sum, after further substitution and simplification, we get:
$\dfrac{n}{2}(2a + (n - 1)d)$
$\dfrac{{30}}{2}(\left( {2 \times 10} \right) + (30 - 1)3)$
$15(\left( {20} \right) + (29)3)$
By further simplifying the equation we get,
$15(107) = 1605$
Hence, the sum of all the two-digit numbers which leave a remainder of 1 when divided by 3 is 1605.
The right answer is option (C).
Note: The above solution is when we know the number of terms, common difference and first term in the sequence even without knowing the last term. We can use an alternative method where we know the number of terms, first term and last term in the sequence $\dfrac{n}{2}(a + {a_n})$ where ${a_n}$ is the last term in the sequence and $a$ is the first term in the sequence.
For an AP, the sum of n terms = $\dfrac{n}{2}(2a + (n - 1)d)$
For a GP, the sum of n terms = $\dfrac{{a({r^n} - 1)}}{{r - 1}}$
For an AGP, the sum of n terms = $\dfrac{{a - [a + (n - 1)d]{r^n}}}{{1 - r}} + \dfrac{{dr(1 - {r^{n - 1}})}}{{{{(1 - r)}^2}}}$
Where
$r$= common ratio
$d$= common difference
$a$= first term in the sequence
$n$= number of terms
Complete step by step answer:
The question asks us to find the sum of all two-digit numbers which leave a remainder 1 when divided by 3, first let’s list out all the numbers that follow the above criteria.
The numbers are: 10, 13, 16, … , 97
The above numbers are in an AP with a common difference of 3. The first term is 10 and the last term is 97.
Now for finding the sum we can use the formula for finding sum of n terms in an AP. a = 10, d = 3
For finding the value of n, we can use the general formula of nth term in an AP.
Which is $a + (n - 1)d$
Now after substituting all the values in the formula, we get:
$10 + (n - 1)3 = 97$
$3(n - 1) = 97 - 10 = 87$$(n - 1) = \dfrac{{87}}{3} = 29$
$n = 29 + 1 = 30$
Now we have all the required values to find the sum, after further substitution and simplification, we get:
$\dfrac{n}{2}(2a + (n - 1)d)$
$\dfrac{{30}}{2}(\left( {2 \times 10} \right) + (30 - 1)3)$
$15(\left( {20} \right) + (29)3)$
By further simplifying the equation we get,
$15(107) = 1605$
Hence, the sum of all the two-digit numbers which leave a remainder of 1 when divided by 3 is 1605.
The right answer is option (C).
Note: The above solution is when we know the number of terms, common difference and first term in the sequence even without knowing the last term. We can use an alternative method where we know the number of terms, first term and last term in the sequence $\dfrac{n}{2}(a + {a_n})$ where ${a_n}$ is the last term in the sequence and $a$ is the first term in the sequence.
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