Question

Find four numbers in A.P. such that the sum of $2$nd and $3$rd terms is $22$ and the product of $1$st and $4$th terms is $85$.

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 $1$st, $2$nd, $3$rd and $4$th term of an A.P are:

$t_1=a+(1-1)d=a$

$t_2=a+(2-1)d=a+d$

$t_3=a+(3-1)d=a+2d$

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

It is given that the sum of $2$nd and $3$rd term is $22$, therefore,

$(a+d)+(a+2d)=22\Rightarrow 2a+3d=22\Rightarrow d=\frac{22-2a}{3}.......\left(1\right)$

Also, it is given that the product of $1$st and $4$th term is $85$, therefore,

$a(a+3d)=85\Rightarrow a(a+3\left(\frac{22-2a}{3}\right))=85\Rightarrow a(a+22-2a)=85\Rightarrow a(-a+22)=85\Rightarrow -a^2+22a=85\Rightarrow a^2-22a+85=0\Rightarrow a^2-17a-5a+85=0\Rightarrow a(a-17)-5(a-17)=0\Rightarrow a-17=0,a-5=0\Rightarrow a=17,a=5$

Let us take $a=5$ and substitute it in equation 1:

$d=\frac{22-(2\times 5)}{3}=\frac{22-10}{3}=\frac{12}{3}=4$

Now, $a_1=5$ and $d=4$, therefore, the next three terms of an A.P are:

$a_2=a_1+d=5+4=9$

$a_3=a_2+d=9+4=13$

$a_4=a_3+d=13+4=17$

Hence, the four terms of an A.P are $5,9,13,17$ and if we take $a=17$, then we get the vice versa A.P. that is $17,13,9,5$