JEE Main 2021 — Mathematics Algebra

Given, $\frac{{a}_{1}+{a}_{2}+\ldots +{a}_{10}}{{a}_{1}+{a}_{2}+\ldots +{a}_{p}}=\frac{{S}_{10}}{{S}_{p}}=\frac{100}{{p}^{2}}$a1​+a2​+…+ap​a1​+a2​+…+a10​​=Sp​S10​​=p2100​

$$\Rightarrow \frac{\frac{10}{2}[2{a}_{1}+9d]}{\frac{p}{2}[2{a}_{1}+(p-1)d]}=\frac{100}{{p}^{2}}$$⇒2p​[2a1​+(p−1)d]210​[2a1​+9d]​=p2100​

$$\Rightarrow \frac{10[2{a}_{1}+9d]}{p[2{a}_{1}+(p-1)d]}=\frac{100}{{p}^{2}}$$⇒p[2a1​+(p−1)d]10[2a1​+9d]​=p2100​

$$\Rightarrow \frac{2{a}_{1}+9d}{2{a}_{1}+(p-1)d}=\frac{10}{p}$$⇒2a1​+(p−1)d2a1​+9d​=p10​

Let, ${a}_{1}=a$a1​=a

$$\Rightarrow 2pa+9pd=20a+10pd-10d$$⇒2pa+9pd=20a+10pd−10d

$$\Rightarrow 2pa-20a=pd-10d$$⇒2pa−20a=pd−10d

$$\Rightarrow 2a(p-10)=d(p-10)$$⇒2a(p−10)=d(p−10)

$$\Rightarrow 2a=d$$⇒2a=d

$$\Rightarrow \frac{a}{d}=\frac{1}{2}$$⇒da​=21​

Now, $\frac{{a}_{11}}{{a}_{10}}=\frac{a+10d}{a+9d}$a10​a11​​=a+9da+10d​

$$\Rightarrow \frac{{a}_{11}}{{a}_{10}}=\frac{d(\frac{a}{d}+10)}{d(\frac{a}{d}+9)}$$⇒a10​a11​​=d(da​+9)d(da​+10)​

$$\Rightarrow \frac{{a}_{11}}{{a}_{10}}=\frac{\frac{1}{2}+10}{\frac{1}{2}+9}$$⇒a10​a11​​=21​+921​+10​

$$\Rightarrow \frac{{a}_{11}}{{a}_{10}}=\frac{21}{19}$$⇒a10​a11​​=1921​