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Sequences and Series question

2024 · 30 Jan · Shift 2 · Q60

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2024 · 30 Jan · Shift 2 · Q60

Sequences and Series question

2024 · 30 Jan · Shift 2 · Q60

JEE MainMathematicsSequences and SeriesNumerical+4 / −1

Let

S_n

be the sum to

n

-terms of an arithmetic progression

3,7,11

, If

40\lt \left(\frac{6}{n(n+1)} \sum_{k=1}^n S_k\right)\lt 42

, then

n

equals

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 9

Given A.P.

The arithmetic progression is $3,7,11,\dots$ with first term $a=3$ and common difference $d=4$ .

Find $S_n$

Sum of first $n$ terms of an A.P. is $S_n=\frac{n}{2}[2a+(n-1)d].$ So,

=\frac{n}{2}[6+4n-4] =\frac{n}{2}(4n+2) =n(2n+1).$$ Thus, $$S_k=k(2k+1)=2k^2+k.$$ 3. **Compute $\sum_{k=1}^n S_k$** $$\sum_{k=1}^n S_k=\sum_{k=1}^n (2k^2+k) =2\sum_{k=1}^n k^2+\sum_{k=1}^n k.$$ Using formulas, $$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}, \qquad \sum_{k=1}^n k=\frac{n(n+1)}{2}.$$ Hence, $$\sum_{k=1}^n S_k =2\cdot \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$$ $$=\frac{n(n+1)(2n+1)}{3}+\frac{n(n+1)}{2}$$ $$=n(n+1)\left(\frac{2n+1}{3}+\frac{1}{2}\right).$$ Take LCM $6$: $$\frac{2(2n+1)+3}{6}=\frac{4n+5}{6}.$$ So, $$\sum_{k=1}^n S_k=\frac{n(n+1)(4n+5)}{6}.$$ 4. **Substitute into the given expression** Given $$40<\left(\frac{6}{n(n+1)}\sum_{k=1}^n S_k\right)<42.$$ Substitute $\sum_{k=1}^n S_k$: $$\frac{6}{n(n+1)}\cdot \frac{n(n+1)(4n+5)}{6}=4n+5.$$ So the inequality becomes $$40<4n+5<42.$$ 5. **Solve the inequality** Subtract $5$ throughout: $$35<4n<37.$$ Divide by $4$: $$\frac{35}{4}<n<\frac{37}{4}$$ $$8.75<n<9.25.$$ Since $n$ is an integer, $$n=9.$$ 6. **Compare with stored answer** Derived answer: $9$ Stored correct answer: $9$ They match.

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