Question

A person buys a bike by way of loan. He clears the loan in 10 increased installments.
Every month he remains the installer. If the sum of ${{3}^{rd}}$ and ${{9}^{th}}$ instalments is Rs.15000 and sum of ${{2}^{nd}}$ and ${{6}^{th}}$ instalments is Rs.13000. find last instalment amount and the total amount remitted by him

Answer 1

Hint: To solve this question, we have to use the equation to find the ${{n}^{th}}$ term of the arithmetic progression \[{{a}_{n}}=a+(n-1)d\]. Also use the equation for finding the sum of the ${{n}^{th}}$ is ${{S}_{n}}=\dfrac{n}{2}(a+{{a}_{n}})$. Here, ‘a’ is the first term, ‘d’ is the common difference.

Complete step by step answer:
We are given that the sum of ${{3}^{rd}}$ and ${{9}^{th}}$ instalments is Rs.15000. So, we can write,
$\begin{align}
  & {{a}_{3}}=a+(3-1)d \\
 & \Rightarrow {{a}_{3}}=a+2d \\
\end{align}$
Similarly, we can write,
$\begin{align}
  & {{a}_{9}}=a+(9-1)d \\
 & \Rightarrow {{a}_{9}}=a+8d \\
\end{align}$
So, we can write,
$\begin{align}
  & {{a}_{3}}+{{a}_{9}}=(a+2d)+(a+8d) \\
 & \Rightarrow {{a}_{3}}+{{a}_{9}}=2a+10d=15000 \\
\end{align}$
Similarly, it is given that the sum of ${{2}^{nd}}$ and ${{6}^{th}}$ instalments is Rs.13000. So,
$\begin{align}
  & {{a}_{2}}=a+(2-1)d \\
 & \Rightarrow {{a}_{2}}=a+d \\
\end{align}$
And
$\begin{align}
  & {{a}_{6}}=a+(6-1)d \\
 & \Rightarrow {{a}_{6}}=a+5d \\
\end{align}$
So, we can write,
$\begin{align}
  & {{a}_{2}}+{{a}_{6}}=(a+d)+(a+5d) \\
 & \Rightarrow {{a}_{2}}+{{a}_{6}}=2a+6d=13000 \\
\end{align}$
So we got two equations. They are,
$2a+10d=15000$…….(1)
$2a+6d=13000$………(2)
Now, we can solve the two equations by subtracting. So, we get,
4d = 2000
$\Rightarrow d=\dfrac{2000}{4}=500$
Therefore, d = 500.
Now, we can substitute the value in equation (1). So, we get,
$2a+10d=15000$
$\Rightarrow $2a + 10(500) = 15000
$\Rightarrow $2a + 5000 = 15000
$\Rightarrow $2a = 10000
$\Rightarrow $a = 5000
Now, we are asked to find the last instalment. So, we have to find ${{a}_{10}}$.
 ${{a}_{10}}$ = a + (10 - 1)d
$\Rightarrow $${{a}_{10}}$ = a + 9d
$\Rightarrow $${{a}_{10}}$ = 5000 + 9(500)
$\Rightarrow $${{a}_{10}}$ = 5000 + 4500
$\Rightarrow $${{a}_{10}}$ = 9500
Therefore, the last instalment is Rs.9500.
Now, we are asked to find the total amount remitted. So, we have to find the sum of the ${{10}^{th}}$ term.
${{S}_{n}}=\dfrac{n}{2}(a+{{a}_{n}})$
$\Rightarrow $${{S}_{10}}=\dfrac{10}{2}(a+{{a}_{10}})$
$\Rightarrow $${{S}_{10}}=\dfrac{10}{2}(5000+9500)$
$\Rightarrow $${{S}_{10}}=5(14500)$
$\Rightarrow $${{S}_{10}}=72500$
Therefore, the total amount remitted is Rs.72500

Note:
There is another equation to find the sum of n terms. That is, ${{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]$. By using this equation also we can find the sum.
${{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]$
$\Rightarrow {{S}_{10}}=\dfrac{10}{2}[2(5000)+(10-1)(4500)]$
$\begin{align}
  & \Rightarrow {{S}_{10}}=5[10000+4500] \\
 & \Rightarrow {{S}_{10}}=5[14500] \\
 & \Rightarrow {{S}_{10}}=72500 \\
\end{align}$
So, by both the equations we get the same answer.