Verified Solution • 2025–26 Session moderate

QJEEAdv22Ma031c: JEE Advanced 2022 Mathematics - Class 11 JEEAdv22Ma031c Solution: Let be an arithmetic progression with and common difference 8.…

Let a1,a2,a3,…a_{1}, a_{2}, a_{3}, \ldots be an arithmetic progression with a1=7a_{1}=7 and common difference 8. Let T1,T2,T3,…T_{1}, T_{2}, T_{3}, \ldots be such that T1=3T_{1}=3 and Tn+1−Tn=anT_{n+1}-T_{n}=a_{n} for n≥1n \geq 1 . Then, which of the following is/are TRUE ?

Multiple Choice Options

  • A T20=1604T_{20}=1604
  • B ∑k=120Tk=10510\sum\limits_{k=1}^{20} T_{k}=10510
  • C T30=3454T_{30}=3454
  • D ∑k=130Tk=35610\sum\limits_{k=1}^{30} T_{k}=35610

Step-by-Step Verified Solution

Here an=7+(n−1)8a_n=7+(n-1) 8 and T1=3\mathrm{T}_1=3 , a1=7,d=8a_1=7, d=8

Also, Tn+1=Tn+an\mathrm{T}_{n+1}=\mathrm{T}_n+a_n

Tn=Tn−1+an−1\mathrm{T}_n=\mathrm{T}_{n-1}+a_{n-1}

T2=T1+a1\mathrm{T}_2=\mathrm{T}_1+a_1

∴Tn+1=(Tn−1+an−1)+an\therefore \mathrm{T}_{n+1}=\left(\mathrm{T}_{n-1}+a_{n-1}\right)+a_n

Tn+1=Tn−2+an−2+an−1+an\mathrm{T}_{n+1}=\mathrm{T}_{n-2}+a_{n-2}+a_{n-1}+a_n

So

amp;Tn+1=T1+a1+a2……anamp; Tn+1=T1+n2(2×7+(n−1)8)amp;Tn+1=T1+n(4n+3)........(1)\begin{aligned}& \mathrm{T}_{n+1}=\mathrm{T}_1+a_1+a_2 \ldots \ldots a_n \\\\& \mathrm{~T}_{n+1}=\mathrm{T}_1+\frac{n}{2}(2 \times 7+(n-1) 8) \\\\& \mathrm{T}_{n+1}=\mathrm{T}_1+n(4 n+3) ........(1)\end{aligned}

For (A), if n=19, T20=3+(19)(79)=1504n=19, \mathrm{~T}_{20}=3+(19)(79)=1504

For (C), if n=29, T30=3+29(119)=3454n=29, \mathrm{~T}_{30}=3+29(119)=3454

For (B),

amp;∑k=120 Tk=∑k=120( T1+4n2+3n)+3amp;=3+∑k=120(3+4n2+3n)amp;=3+3(19)+3(19)(20)2+4(19)(20)(34)6amp;=3+10507=10510\begin{aligned}& \sum_{k=1}^{20} \mathrm{~T}_k =\sum_{k=1}^{20}\left(\mathrm{~T}_1+4 n^2+3 n\right)+3 \\\\& = 3+\sum_{k=1}^{20}\left(3+4 n^2+3 n\right) \\\\&= 3+3(19)+\frac{3(19)(20)}{2} +\frac{4(19)(20)(34)}{6} \\\\&= 3+10507=10510\end{aligned}

Similarly, for (D)

∑k=130 Tk=3+∑k=129(4n2+3n+3)=35615\sum\limits_{k=1}^{30} \mathrm{~T}_k=3+\sum\limits_{k=1}^{29}\left(4 n^2+3 n+3\right)=35615

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