JEE Main 2019 — Mathematics Algebra

We know that, the ${n}^{th}$nth term of a $G.P.$G.P. with first term $a$a and common ratio $r$r is ${t}_{n}=a{r}^{n-1}.$tn​=arn−1.

Then, $b=ar$b=ar and $c=a{r}^{2}$c=ar2

Now, $3a,7b$3a,7b and $15c$15c are in $A.P.$A.P.

$$\Rightarrow 14b=3a+15c$$⇒14b=3a+15c

$$\Rightarrow 14(ar)=3a+15(a{r}^{2})$$⇒14(ar)=3a+15(ar2)

$$\Rightarrow 14r=3+15{r}^{2}$$⇒14r=3+15r2

$$\Rightarrow 15{r}^{2}-14r+3=0$$⇒15r2−14r+3=0

$$\Rightarrow (3r-1)(5r-3)=0$$⇒(3r−1)(5r−3)=0

$\Rightarrow r=\frac{1}{3}$⇒r=31​ or $r=\frac{3}{5}$r=53​

The only acceptable value is $r=\frac{1}{3},$r=31​, because $r\in (0,\frac{1}{2}]$r∈(0,21​]

$\therefore$∴ the first term of the $A.P.$A.P. is $A=3a$A=3a and the common difference is $d=7b-3a=7ar-3a=\frac{7}{3}a-3a=-\frac{2}{3}a$d=7b−3a=7ar−3a=37​a−3a=−32​a

Also, we know that the ${n}^{th}$nth term of an $A.P.$A.P. with first term $A$A and common difference $d$d is ${T}_{n}=A+(n-1)d$Tn​=A+(n−1)d

$$\therefore {T}_{4}=3a+(4-1)(-\frac{2}{3}a)= 3a-2a=a.$$∴T4​=3a+(4−1)(−32​a)=3a−2a=a.