Verified Solution • 2025–26 Session moderate

QJEEM24Ma5859: JEE Main 2024 Mathematics - Class 11 JEEM24Ma5859 Solution: If each term of a geometric progression with and , is the…

If each term of a geometric progression a1,a2,a3,…a_1, a_2, a_3, \ldots with a1=18a_1=\frac{1}{8} and a2≠a1a_2 \neq a_1 , is the arithmetic mean of the next two terms and Sn=a1+a2+…..+anS_n=a_1+a_2+\ldots . .+a_n , then S20−S18S_{20}-S_{18} is equal to

Multiple Choice Options

  • A −215-2^{15}
  • B 2152^{15}
  • C −218-2^{18}
  • D 2182^{18}

Step-by-Step Verified Solution

Let r′r^{\prime} th term of the GP be an−1a^{n-1} . Given,

amp;2ar=ar+1+ar+2amp;2arn−1=arn+arn+1amp;2r=1+ramp;r2+r−2=0\begin{aligned}& 2 a_r=a_{r+1}+a_{r+2} \\& 2 a r^{n-1}=a r^n+a r^{n+1} \\& \frac{2}{r}=1+r \\& r^2+r-2=0\end{aligned}

Hence, we get, r=−2r=-2 (as r≠1r \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=ar18(1+r)\mathrm{~T}_{19}+\mathrm{T}_{20}=\mathrm{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}

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