Let a _ 1 , a _ 2 , a _ 3 , be an arithmetic progression with a _ 1 = 7 and common difference 8 . Let T _ 1 , T _ 2 , T _ 3 , be such that T _ 1 = 3 and T _ n + 1 - T _ n = a _ n fo n ≥ 1 . Then, which of the following

Correct answer: B. ∑ _ k = 1 ^ 20 T _ k = 10510 Given, a _ n = 7 + (n - 1)8 and T _ 1 = 3 Also T _ n + 1 = T _ n + a _ n Now T _ n = T _ n - 1 + a _ n - 1 Putting n = 1 we get, T _ 2 = T _ 1 + a _ 1 Now putting n = 2 we get, T _ 3 = T _ 2 + a _ 2 ⇒ T _ 3 = T _ 1 + a _ 1 + a _ 2 And so on we get, ⇒ T _ n + 1 = T _ 1 + a _ 1 + a _ 2 + + a _ n ⇒ T _ n + 1 = T _ 1 + (n)/(2)[2(7) + (n - 1)8] ⇒ T _ n + 1 = T _ 1 + n(4n + 3) ⋯ (1) Now for n = 19 we get, T _ 20 = 3 + (19)(79) = 1504 And for n = 29 we get, T _ 30 = 3 + (29)(119) = 3454 option C is correct Now finding sum we get, ∑ _ k = 1 ^ 20 T _ k = 3 + ∑ _ k = 2 ^ 20 T _ k = 3 + ∑ _ k = 1 ^ 19 (3 + 4 n ^ 2 + 3n) = 3 + 3(19) + (3(19)(20))/(2) + (4(19)(20)(39))/(6) = 3 + 10507 = 10510 option B is correct And Similarly ∑ _ k = 1 ^ 30 T _ k = 3 + ∑ _ k = 1 ^ 29 (4 n ^ 2 + 3n + 3) = 35615

JEE Advanced 2022Paper 1

Exam: JEE Advanced
Year: 2022
Paper: Paper 1
Subject: Mathematics
Chapter: Sequences and Series

JEE Advanced 2022 · Paper 1 · Mathematics · Sequences and Series

Let a _ 1 , a _ 2 , a _ 3 , be an arithmetic progression with a _ 1...

Let

{a}_{1}, {a}_{2}, {a}_{3}, \ldots

be an arithmetic progression with

{a}_{1} = 7

and common difference

8

. Let

{T}_{1}, {T}_{2}, {T}_{3}, \ldots

be such that

{T}_{1} = 3

and

{T}_{n + 1} - {T}_{n} = {a}_{n}

n \ge 1

. Then, which of the following is/are TRUE?

Options

A ${T}_{20} = 1604$
B $\mathop{∑}_{k = 1}^{20}{T}_{k} = 10510$
C ${T}_{30} = 3454$
D $\mathop{∑}_{k = 1}^{30}{T}_{k} = 35610$

Correct answer

B. $\mathop{∑}_{k = 1}^{20}{T}_{k} = 10510$

Step-by-step solution

Given,

${a}_{n} = 7 + \left(n - 1\right)8 and$ {T}_{1} = 3 $Also$ {T}_{n + 1} = {T}_{n} + {a}_{n} $Now$ {T}_{n} = {T}_{n - 1} + {a}_{n - 1} $Putting$ n = 1 $we get, $ {T}_{2} = {T}_{1} + {a}_{1} $Now putting$ n = 2 $we get,$ {T}_{3} = {T}_{2} + {a}_{2} ⇒ {T}_{3} = {T}_{1} + {a}_{1} + {a}_{2} $And so on we get,$ ⇒ {T}_{n + 1} = {T}_{1} + {a}_{1} + {a}_{2} + \ldots + {a}_{n} $$ ⇒ {T}_{n + 1} = {T}_{1} + \dfrac{n}{2}\left[2\left(7\right) + \left(n - 1\right)8\right] $$ ⇒ {T}_{n + 1} = {T}_{1} + n\left(4n + 3\right) ⋯ \left(1\right) $Now for$ n = 19 $we get,$ {T}_{20} = 3 + \left(19\right)\left(79\right) = 1504 $And for$ n = 29 $we get,$ {T}_{30} = 3 + \left(29\right)\left(119\right) = 3454 ${option C is correct}Now finding sum we get,$ \mathop{∑}_{k = 1}^{20}{T}_{k} = 3 + \mathop{∑}_{k = 2}^{20}{T}_{k} = 3 + \mathop{∑}_{k = 1}^{19}\left(3 + 4{n}^{2} + 3n\right) $$ = 3 + 3\left(19\right) + \dfrac{3\left(19\right)\left(20\right)}{2} + \dfrac{4\left(19\right)\left(20\right)\left(39\right)}{6} $$ = 3 + 10507 = 10510 ${option B is correct}And Similarly$ \mathop{∑}_{k = 1}^{30}{T}_{k} = 3 + \mathop{∑}_{k = 1}^{29}\left(4{n}^{2} + 3n + 3\right) = 35615$

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