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Sequences and Series question

2022 · 24 Jun · Shift 1 · Q31
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  5. /2022 · 24 Jun · Shift 1 · Q31

Sequences and Series question

2022 · 24 Jun · Shift 1 · Q31

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
If {ai}i=1n\{ {a_i}\} _{i = 1}^n{ai​}i=1n​, where n is an even integer, is an arithmetic progression with common difference 1, and ∑i=1nai=192, ∑i=1n/2a2i=120\sum\limits_{i = 1}^n {{a_i} = 192} ,\,\sum\limits_{i = 1}^{n/2} {{a_{2i}} = 120}i=1∑n​ai​=192,i=1∑n/2​a2i​=120, then n is equal to :
  1. A
    48
  2. B
    96
  3. C
    92
  4. D
    104
View written solutionFree

Correct answer: B

  1. Let the arithmetic progression be

    \qquad a_i=a+(i-1),$$ since the common difference is $1$.

  2. Use the sum of all nnn terms: ∑i=1nai=n2(2a+(n−1))=192.\sum_{i=1}^n a_i=\frac{n}{2}\big(2a+(n-1)\big)=192.∑i=1n​ai​=2n​(2a+(n−1))=192. So, n(2a+n−1)=384.(1)n(2a+n-1)=384. \qquad (1)n(2a+n−1)=384.(1)

  3. Now consider the sum of even-indexed terms: a2i=a+(2i−1),i=1,2,…,n2.a_{2i}=a+(2i-1), \qquad i=1,2,\dots,\frac n2.a2i​=a+(2i−1),i=1,2,…,2n​. Thus a2,a4,…,ana_2,a_4,\dots,a_na2​,a4​,…,an​ itself forms an AP with:

    • first term a2=a+1a_2=a+1a2​=a+1
    • common difference 222
    • number of terms n2\frac n22n​

    Its sum is given as 120120120: ∑i=1n/2a2i=n/22(2(a+1)+(n2−1)2)=120.\sum_{i=1}^{n/2} a_{2i}=\frac{n/2}{2}\Big(2(a+1)+\Big(\frac n2-1\Big)2\Big)=120.∑i=1n/2​a2i​=2n/2​(2(a+1)+(2n​−1)2)=120.

    Simplify inside: 2(a+1)+(n2−1)2=2a+2+n−2=2a+n.2(a+1)+\Big(\frac n2-1\Big)2=2a+2+n-2=2a+n.2(a+1)+(2n​−1)2=2a+2+n−2=2a+n. Hence, n4(2a+n)=120,\frac{n}{4}(2a+n)=120,4n​(2a+n)=120, so n(2a+n)=480.(2)n(2a+n)=480. \qquad (2)n(2a+n)=480.(2)

  4. Subtract equation (1) from equation (2): n(2a+n)−n(2a+n−1)=480−384,n(2a+n)-n(2a+n-1)=480-384,n(2a+n)−n(2a+n−1)=480−384, n=96.n=96.n=96.

  5. Therefore, n=96.\boxed{n=96}.n=96​.

  6. Checking options:

    • A: 484848 ❌
    • B: 969696 ✅
    • C: 929292 ❌
    • D: 104104104 ❌

So the correct option is B.

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