Question

In an arithmetic series, the sum of first $11$ terms is $44$ and that of the next $11$ terms is $55$. Find the arithmetic series

Answer 1

It is given that the sum of first $11$ terms is $44$ that is $S_{11}=44$.

We know that the sum of an arithmetic series with first term $a$ and common difference $d$ is $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$, therefore,

$S_{11}=\frac{11}{2}[2a+(11-1\left)d\right]\Rightarrow 44\times 2=11(2a+10d)\Rightarrow 88=11(2a+10d)\Rightarrow \frac{88}{11}=2a+10d\Rightarrow 2a+10d=\mathrm{8......}\left(1\right)$

It is also given that the sum of next $11$ terms is $55$ that is

$S_{22}=S_{11}+55\Rightarrow S_{22}=44+55(\because S_{11}=44)\Rightarrow S_{22}=99$

Now the sum of first $22$ terms is $99$, therefore,

$S_{22}=\frac{22}{2}[2a+(22-1\left)d\right]\Rightarrow 99\times 2=22(2a+21d)\Rightarrow 198=22(2a+21d)\Rightarrow \frac{198}{22}=2a+21d\Rightarrow 2a+21d=\mathrm{9......}\left(2\right)$

Subtract equation 1 from equation 2 as follows:

$(2a-2a)+(21d-10d)=9-8\Rightarrow 11d=1\Rightarrow d=\frac{1}{11}$

Now, substitute the value of $d$ in equation 1:

$2a+(10\times \frac{1}{11})=8\Rightarrow 2a+\frac{10}{11}=8\Rightarrow 22a+10=88\Rightarrow 22a=88-10\Rightarrow 22a=78\Rightarrow a=\frac{78}{22}\Rightarrow a=\frac{39}{11}$

Therefore, the terms of the arithmetic series are:

$a_1=\frac{39}{11}$

$a_2=a_1+d=\frac{39}{11}+\frac{1}{11}=\frac{40}{11}$

$a_3=a_2+d=\frac{40}{11}+\frac{1}{11}=\frac{41}{11}$

Hence, the required arithmetic series is $\frac{39}{11}+\frac{40}{11}+\frac{41}{11}+......$