Question

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

Answer 1

We know that the formula for the nth term is $t_n=a+(n-1)d$, where $a$ is the first term, $d$ is the common difference.

The $4$th, $6$th,$8$th and $10$th term of an A.P are:

$t_4=a+(4-1)d=a+3d$

$t_6=a+(6-1)d=a+5d$

$t_8=a+(8-1)d=a+7d$

$t_{10}=a+(10-1)d=a+9d$

It is given that the sum of $4$th and $8$th term is $24$, therefore,

$(a+3d)+(a+7d)=24\Rightarrow 2a+10d=24\Rightarrow a+5d=12......\left(1\right)$

Also, it is given that the sum of $6$th and $10$th term is $44$, therefore,

$(a+5d)+(a+9d)=44\Rightarrow 2a+14d=44\Rightarrow a+7d=22......\left(2\right)$

Now, subtract equation 1 from 2 as follows:

$(a-a)+(7d-5d)=22-12\Rightarrow 2d=10\Rightarrow d=\frac{10}{2}=5$

Substitute the value of the difference $d=5$ in equation 1:

$a+(5\times 5)=12\Rightarrow a+25=12\Rightarrow a=12-25=-13$

Now, the first term is $a_1=-13$ and the difference is $d=5$, therefore, the next two terms of the A.P are:

$a_2=a_1+d=-13+5=-8$

$a_3=a_2+d=-8+5=-3$

Hence, the first three terms are $-13,-8,-3$.