Question

The sum of $4\mathrm{th}$and $8\mathrm{th}$ terms of an A.P. is $24$ and the sum of the $6\mathrm{th}$ and $10\mathrm{th}$ terms is $44$. Find the first three terms of the A.P.

Answer 1

Step 1: Determine the integers

The $\mathrm{nth}$ term the AP (arithmetic progression) is given by the for

$\mathrm{an}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}$

where; $\mathrm{a}=$ first term of AP

and, $\mathrm{d}=$ a common difference between AP

and, $\mathrm{an}=$$\mathrm{nth}$ term of AP

We know that, $\mathrm{an}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}$.

Therefore,

$4\mathrm{th}$ term is given by: $\mathrm{a}4=\mathrm{a}+(4-1)\mathrm{d}=\mathrm{a}+3\mathrm{d}$

$8\mathrm{th}$ term is given by: $\mathrm{a}8=\mathrm{a}+(8-1)\mathrm{d}=\mathrm{a}+7\mathrm{d}$

$6\mathrm{th}$ term is given by $\mathrm{a}6=\mathrm{a}+(6-1)\mathrm{d}=\mathrm{a}+5\mathrm{d}$

$10\mathrm{th}$ term is given by $\mathrm{a}10=\mathrm{a}+(10-1)=\mathrm{a}+9\mathrm{d}$

Step 2: Find the terms

It is given that, Sum of $4\mathrm{th}\&8\mathrm{th}$ terms of AP is $24$
Which implies,$\mathrm{a}4+98=24$
Putting values we get:
$\Rightarrow$$\mathrm{a}+3\mathrm{d}+\mathrm{a}+7\mathrm{d}=24$

$\Rightarrow$ $2\mathrm{a}+10\mathrm{d}=24$
Now we will divide by $2$ both sides, we get:
$\Rightarrow$ $\mathrm{a}+5\mathrm{d}=12$$....\left(1\right)$
It is also given that the sum of $6\mathrm{th}$ and $10\mathrm{th}$ term of AP is $44$
$\Rightarrow$ $\mathrm{a}6+\mathrm{a}10=44$

Putting values we get,

$\Rightarrow$$\mathrm{a}+5\mathrm{d}+\mathrm{a}+9\mathrm{d}=44$
Simplifying further,
$\Rightarrow$ $2\mathrm{a}+14\mathrm{d}=44$
Dividing by $2$ on both sides, we get
$\Rightarrow$ $\mathrm{a}+7\mathrm{d}=22$ $....\left(2\right)$

Step 3: Determine the common difference

Solving equations $\left(1\right)$ and $\left(2\right)$

we get,$-2\mathrm{d}=-10$

which further simplifies to,
$\Rightarrow$ $\text{d}={\displaystyle \frac{-10}{-2}}=5$
Therefore, $\mathrm{d}=5$ or common difference $=5$

Step 4: Determine the terms of integers

From $\left(1\right)$,we have: $\mathrm{a}+5\mathrm{d}=12$
Putting the values of d we get,

$\Rightarrow$ $\mathrm{a}=12-5\mathrm{d}$
$\Rightarrow$ $=12-25$
$\Rightarrow$$\mathrm{a}=-13$

First term of AP, $\mathrm{a}=-13$
Or, $\mathrm{a}1=-13$
Second term of AP;

$\Rightarrow$$\mathrm{a}2=\mathrm{a}+(2-1)\mathrm{d}$
$\Rightarrow$ $=\mathrm{a}+\mathrm{d}$
$\Rightarrow$ $=-13+5$
$\Rightarrow$ $=-8$
Third term of AP;

$\Rightarrow$$\mathrm{a}3=\mathrm{a}+(3-1)\mathrm{d}$
$\Rightarrow$ $=\mathrm{a}+2\mathrm{d}$
$\Rightarrow$ $=-13+2\times 5$
$\Rightarrow$ $=-13+10$
$\Rightarrow$ $=-3$
$\therefore \mathrm{a}1=-13,\mathrm{a}2=-8,\mathrm{a}3=-3$

Hence, the three integers are $-13,-8,-3$.