Question

A manufacture of radio sets produced $600$ units in the third year and $700$ units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find
$i)$ The production in the first year
$ii)$ The total products in $7$ years and
$iii)$ The produce in the ${10}^{th}$ year.

Answer 1

Let $a_n$ denotes the Prodution of radio sets in the $n^{th}$ year
Let $a$ be the production of the first year
Let $d$ be the common difference.
The $n^{th}$ term of an A.P can be expressed as:
$a_n=a+(n-1)d$
Part $\left(i\right)$: Given: $a_3=600$
$\Rightarrow a+2d=600\dots \left(i\right)$
Also, $a_7=700$
$\Rightarrow a+6d=700\dots \left(ii\right)$
Subtracting equation $\left(i\right)$ from equation $\left(ii\right)$, we get:
$\Rightarrow 4d=100\Rightarrow d=25$
Putting the value of $d$ in the equation $\left(i\right)$, we get:
$\Rightarrow a+50=600\therefore a=550$
Therfore, the production in the first year is $550$ units.

Part$\left(ii\right)$: The sum of $n$ term of an A.P can be expressed as:
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
For $S_7$
$\Rightarrow s_7=\frac{7}{2}[2\times 550+(7-1\left)25\right]$
$\Rightarrow s_7=\frac{7}{2}[1100+150]$
$\Rightarrow s_7=4375\text{units}$
Therefore, the total product in the $7$ year is $4375\text{units}$

Part $\left(iii\right)$:
We know, $a_n=a+(n-1)d\Rightarrow a_{10}=a+(10-1)d\Rightarrow a_{10}=550+9\left(25\right)\Rightarrow a_{10}=775$
Therefore, the total product in the ${10}^{th}$ year is $775\text{units}$