Answer
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Hint: To solve this question, we need to know the basic concepts of arithmetic progression. Here, the first term of the given arithmetic progression is the smallest integer greater than 10 which is a multiple of 4, while the last term would be the largest integer just smaller than 250 which is a multiple of 4. We need to form an arithmetic progression with a common difference as 4. The formula of the ${{n}^{th}}$ term of this series is given by -
${{a}_{n}}$= a + (n-1) d
${{a}_{n}}$= ${{n}^{th}}$ term of this series
a = first term of the series
n = number of terms in the series
d= common difference (4 in this case)
The value of n will give the required answer.
Complete step-by-step solution -
Since, we need to find the arithmetic progression to solve the problem, we first find the first term of the arithmetic progression. Thus, we first start by finding the smallest integer greater than 10 which is a multiple of 4. This is clearly 12 (since, 11 is not multiple of 4 and thus 12 would be the required number). Now, we find the last term. For this we start with 249 (largest number just smaller than 250). Since, this is not a multiple of 4, we then move onto 248. This is in fact a multiple of 4, thus we have found our last term. Since, we have to find the multiple of 4, the common difference is 4. Thus, we have in the equation ${{a}_{n}}$= a + (n-1) d, we have,
${{a}_{n}}$ = 248
a = 12
d = 4
Now, substituting the values, we have,
248 = 12 + 4 (n-1)
(n-1) = $\dfrac{248-12}{4}$
(n-1) = 59
n = 60
Thus, the number of multiples of 4 that lie in between 10 and 250 are 60.
Note: Another way to find the answer between two numbers (say a and b) is by doing the following. We first find the remainder when we divide 250 by 4 (which is 2). We then subtract 2 from 250 to get the last term (250 – 2 = 248) of the arithmetic progression. We then find the remainder when we divide 10 by 4 (which is 2). We then add 2 to 10 (which is 10 + 2=12) to get the first term. Now, we find the answer by using the formula –
$\dfrac{248}{4}-\dfrac{12}{4}+1$ = 62 – 3 + 1 = 60 (which is the same answer as the one in the solutions.)
${{a}_{n}}$= a + (n-1) d
${{a}_{n}}$= ${{n}^{th}}$ term of this series
a = first term of the series
n = number of terms in the series
d= common difference (4 in this case)
The value of n will give the required answer.
Complete step-by-step solution -
Since, we need to find the arithmetic progression to solve the problem, we first find the first term of the arithmetic progression. Thus, we first start by finding the smallest integer greater than 10 which is a multiple of 4. This is clearly 12 (since, 11 is not multiple of 4 and thus 12 would be the required number). Now, we find the last term. For this we start with 249 (largest number just smaller than 250). Since, this is not a multiple of 4, we then move onto 248. This is in fact a multiple of 4, thus we have found our last term. Since, we have to find the multiple of 4, the common difference is 4. Thus, we have in the equation ${{a}_{n}}$= a + (n-1) d, we have,
${{a}_{n}}$ = 248
a = 12
d = 4
Now, substituting the values, we have,
248 = 12 + 4 (n-1)
(n-1) = $\dfrac{248-12}{4}$
(n-1) = 59
n = 60
Thus, the number of multiples of 4 that lie in between 10 and 250 are 60.
Note: Another way to find the answer between two numbers (say a and b) is by doing the following. We first find the remainder when we divide 250 by 4 (which is 2). We then subtract 2 from 250 to get the last term (250 – 2 = 248) of the arithmetic progression. We then find the remainder when we divide 10 by 4 (which is 2). We then add 2 to 10 (which is 10 + 2=12) to get the first term. Now, we find the answer by using the formula –
$\dfrac{248}{4}-\dfrac{12}{4}+1$ = 62 – 3 + 1 = 60 (which is the same answer as the one in the solutions.)
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