Question

The sum of the first 10 terms of an arithmetic series is 100 and the sum of next 10 300 .Find the series.​

Answer 1

Let a be the first term in the sequence, and d the common difference between consecutive terms. If aₙ denotes the n-th term in the sequence, then

a₁ = a

a₂ = a₁ + d = a + d

a₃ = a₂ + d = a + 2d

a₄ = a₃ + d = a + 3d

and so on, up to the n-th term

aₙ = a + (n - 1) d

The sum of the first 10 terms is 100, and so

$\displaystyle \sum_{n=1}^{10} a_n = 100 \\ \sum_{n=1}^{10} (a + (n-1)d) = 100 \\ (a-d) \sum_{n=1}^{10} 1 + d \sum_{n=1}^{10} n = 100 \\ 10a+45d = 100$

where we use the well-known sum formulas,

$\displaystyle \sum_{n=1}^N 1 = 1 + 1 + 1 + \cdots + 1 = N$

$\displaystyle \sum_{n=1}^N n = 1 + 2 + 3 + \cdots + N = \frac{N(N+1)}2$

The sum of the next 10 terms is 300, so

$\displaystyle \sum_{n=11}^{20} a_n = 300 \\ (a-d) \sum_{n=11}^{20} 1 + d \sum_{n=11}^{20} n = 300 \\ (a-d) \left(\sum_{n=1}^{20} 1 - \sum_{n=1}^{10} 1\right) 1 + d \left(\sum_{n=1}^{20} n - \sum_{n=1}^{10} n\right) = 300 \\ 10a+145d = 300$

Solve for a and d. Eliminating a gives

(10a + 145d) - (10a + 45d) = 300 - 100

100d = 200

d = 2

and solving for a gives

10a + 145×2 = 300

10a = 10

a = 1

So, the given sequence is simply the sequence of positive odd integers,

{1, 3, 5, 7, 9, …}

given recursively by the relation

$\begin{cases}a_1 = 1 \\ a_n = a_{n-1} + 2 & \text{for }n>1\end{cases}$

and explicitly by

$a_n = 1 + 2(n-1) = 2n - 1$

for n ≥ 1.