JEE Main 2023 — Mathematics Algebra

Given:

$${T}_{4}=500$$T4​=500

Let $a=$a= first term

$r=$r=common ratio $=\frac{1}{m},m\in N$=m1​,m∈N

So,

$$a{r}^{3}=500$$ar3=500

$$\Rightarrow \frac{a}{{m}^{3}}=500$$⇒m3a​=500

Now,

$${S}_{n}-{S}_{n-1}=a{r}^{n-1}$$Sn​−Sn−1​=arn−1

$${S}_{6}>{S}_{5}+1$$S6​>S5​+1

$$\Rightarrow {S}_{6}-{S}_{5}>1$$⇒S6​−S5​>1

$$\Rightarrow \frac{a}{{m}^{5}}>1$$⇒m5a​>1

$$\Rightarrow \frac{a}{{m}^{3}}\times \frac{1}{{m}^{2}}>1$$⇒m3a​×m21​>1

$$\Rightarrow \frac{500}{{m}^{2}}>1$$⇒m2500​>1

$$\Rightarrow {m}^{2}<500...(1)$$⇒m2<500...(1)

And,

$${S}_{7}-{S}_{6}<\frac{1}{2}$$S7​−S6​<21​

$$\Rightarrow \frac{a}{{m}^{6}}<\frac{1}{2}$$⇒m6a​<21​

$$\Rightarrow \frac{a}{{m}^{3}}\times \frac{2}{{m}^{3}}<1$$⇒m3a​×m32​<1

$$\Rightarrow 500\times \frac{2}{{m}^{3}}<1$$⇒500×m32​<1

$$\Rightarrow {m}^{3}>{10}^{3}$$⇒m3>103

$$\Rightarrow m>10....(2)$$⇒m>10....(2)

From $(1)$(1) and $(2)$(2), we have

$$m=11,12,13\ldots \ldots \ldots \ldots ..,22$$m=11,12,13…………..,22

So, number of possible values of $m$m is $12$12.