# The sum of 15 terms of an arithmetic progression is 600, and the common difference is 5. Find the first term.

Last updated date: 19th Mar 2023

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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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