Answer
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Hint: To find the first term of the given arithmetic progression, we need to form an arithmetic progression with common difference as 5. 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 (5 in this case)
Further, the sum of the series of arithmetic progression is given by –
Sum (of n terms) = $\dfrac{n}{2}$[a+${{a}_{n}}$] = $\dfrac{n}{2}$[2a + (n-1) d]
Complete step by step answer:
To solve this problem, we first start with the equation of sum of n terms since we are given that sum of n terms is given by –
$\dfrac{n}{2}$[2a + (n-1) d]
Here, we have n = 15 and d = 5. Since, this is equal to 600. Thus, we have,
$\dfrac{n}{2}$[2a + (n-1) d] = 600
$\dfrac{15}{2}$[2a + (15-1)$\times $5] = 600
$\dfrac{15}{2}$[2a + (14$\times $5)] = 600
[2a + (14$\times $5)] = $\dfrac{600\times 2}{15}$
[2a + (14$\times $5)] = 80
2a + 70 = 80
2a = 10
a = 5
Hence, the first term is 5.
Note: One of the ways to get the expression of the sum of an arithmetic progression is by doing the following-
= a + (a+d) + (a+2d) + … + a+(n-1) d
We club first and last, second and second last terms and so on. Thus,
= [{a + a + (n-1) d} + {a+d + a +(n-2) d} + …]
= [{2a + (n-1) d} + {2a + (n-1) d} + …]
Since, originally there were n terms and now we have clubbed 2 terms at a time, we are left
with total $\dfrac{n}{2}$terms of {2a + (n-1) d}. Thus, we have,
= $\dfrac{n}{2}${2a + (n-1) d}
Which is the formula of sum of arithmetic progression, which can be used to solve the problem.
${{a}_{n}}$= 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 (5 in this case)
Further, the sum of the series of arithmetic progression is given by –
Sum (of n terms) = $\dfrac{n}{2}$[a+${{a}_{n}}$] = $\dfrac{n}{2}$[2a + (n-1) d]
Complete step by step answer:
To solve this problem, we first start with the equation of sum of n terms since we are given that sum of n terms is given by –
$\dfrac{n}{2}$[2a + (n-1) d]
Here, we have n = 15 and d = 5. Since, this is equal to 600. Thus, we have,
$\dfrac{n}{2}$[2a + (n-1) d] = 600
$\dfrac{15}{2}$[2a + (15-1)$\times $5] = 600
$\dfrac{15}{2}$[2a + (14$\times $5)] = 600
[2a + (14$\times $5)] = $\dfrac{600\times 2}{15}$
[2a + (14$\times $5)] = 80
2a + 70 = 80
2a = 10
a = 5
Hence, the first term is 5.
Note: One of the ways to get the expression of the sum of an arithmetic progression is by doing the following-
= a + (a+d) + (a+2d) + … + a+(n-1) d
We club first and last, second and second last terms and so on. Thus,
= [{a + a + (n-1) d} + {a+d + a +(n-2) d} + …]
= [{2a + (n-1) d} + {2a + (n-1) d} + …]
Since, originally there were n terms and now we have clubbed 2 terms at a time, we are left
with total $\dfrac{n}{2}$terms of {2a + (n-1) d}. Thus, we have,
= $\dfrac{n}{2}${2a + (n-1) d}
Which is the formula of sum of arithmetic progression, which can be used to solve the problem.
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