JEE Main 2024 — Mathematics Algebra

Given, $\mathrm{AP}:3,7,11,15,...$AP:3,7,11,15,...

$$\Rightarrow {a}_{n}=3+(n-1)(4)$$⇒an​=3+(n−1)(4)

$$\Rightarrow {a}_{n}=4n-1$$⇒an​=4n−1

$$\Rightarrow {S}_{n}=\frac{n}{2}[6+4n-4]$$⇒Sn​=2n​[6+4n−4]

$$\Rightarrow {S}_{n}=\frac{n\times 2(2n+1)}{2}$$⇒Sn​=2n×2(2n+1)​

$$\Rightarrow {S}_{n}=2{n}^{2}+n$$⇒Sn​=2n2+n

Now, solving

$$\Rightarrow \sum _{k=1}^{n}{S}_{k}=2\sum _{n=1}^{n}{n}^{2}+\sum _{n=1}^{n}n$$⇒k=1∑n​Sk​=2n=1∑n​n2+n=1∑n​n

$$\Rightarrow \sum _{k=1}^{n}{S}_{k}=\frac{n(n+1)(2n+1)}{3}+\frac{n(n+1)}{2}$$⇒k=1∑n​Sk​=3n(n+1)(2n+1)​+2n(n+1)​

$$\Rightarrow \sum _{k=1}^{n}{S}_{k}=\frac{n(n+1)}{2}[1+\frac{4n+2}{3}]$$⇒k=1∑n​Sk​=2n(n+1)​[1+34n+2​]

Now, using $\displaystyle 40<\frac{6}{n(n+1)}\sum _{k=1}^{n}{S}_{k}<42$40<n(n+1)6​k=1∑n​Sk​<42 we get,

$$\Rightarrow 40<\frac{6}{n(n+1)}\times \frac{n(n+1)}{2}[\frac{5+4n}{3}]<42$$⇒40<n(n+1)6​×2n(n+1)​[35+4n​]<42

$$\Rightarrow 40<4n+5<42$$⇒40<4n+5<42

$$\Rightarrow 35<4n<37$$⇒35<4n<37

$$\Rightarrow n=9$$⇒n=9