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Mathematics · Sequence and Series

JEE Main 2024 — 29 January, Shift 2 — Question 13

If each term of a geometric progression a1,a2,a3,…a_{1}, a_{2}, a_{3}, \ldots with a1=18\mathrm{a}_{1}=\frac{1}{8} and a2≠a1\mathrm{a}_{2} \neq \mathrm{a}_{1}, is the arithmetic mean of the next two terms and Sn=a1+a2+…+an\mathrm{S}_{\mathrm{n}}=\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots+\mathrm{a}_{\mathrm{n}}, then S20−S18\mathrm{S}_{20}-\mathrm{S}_{18} is equal to

  1. Option A:

    2152^{15}

  2. Option B:

    −218-2^{18}

  3. Option C:

    2182^{18}

  4. Option D:

    −215-2^{15}

    Correct

Answer: D

Step-by-step solution

Let r′r^{\prime} th term of the GP be arn−1\mathrm{ar}^{\mathrm{n}-1}. Given, 2ar=ar+1+ar+2,\begin{aligned}&2\mathrm{a}_{\mathrm{r}}=\mathrm{a}_{\mathrm{r}+1}+\mathrm{a}_{\mathrm{r}+2}, & \end{aligned}

2arn−1=arn+arn+1 2 \mathrm{ar}^{\mathrm{n}-1}=\mathrm{ar}^{\mathrm{n}}+\mathrm{ar}^{\mathrm{n}+1}

2r=1+r\frac{2}{\mathrm{r}}=1+\mathrm{r}

r2+r−2=0\mathrm{r}^{2}+\mathrm{r}-2=0

Hence, we get, r=−2(r=-2( as r≠1)r \neq 1)

So, S20−S18=\mathrm{S}_{20}-\mathrm{S}_{18}= (Sum upto 20 terms) - (Sum upto 18 terms) =T19+T20=\mathrm{T}_{19}+\mathrm{T}_{20} T19+T20=ar⁡18(1+r)\mathrm{T}_{19}+\mathrm{T}_{20}=\operatorname{ar}^{18}(1+\mathrm{r})

Putting the values a=18\mathrm{a}=\frac{1}{8} and r=−2\mathrm{r}=-2;

we get T19+T20=−215T_{19}+T_{20}=-2^{15}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 2
Subject
Mathematics
Chapter
Sequence and Series
Topic
Geometric Progression

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