Verified Solution • 2025–26 Session moderate

QJEEM22Ma3ce6: JEE Main 2022 Mathematics - Class 11 JEEM22Ma3ce6 Solution: Let the sum of an infinite G.P., whose first term is a and the…

Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be 9825\frac{98}{25} . Then the sum of the first 21 terms of an AP, whose first term is 10ar,nth 10\mathrm{a r}, \mathrm{n}^{\text {th }} term is an\mathrm{a}_{\mathrm{n}} and the common difference is 10ar210 \mathrm{ar}^{2} , is equal to :

Multiple Choice Options

  • A 21 a1121 \,\mathrm{a}_{11}
  • B 22 a1122 \,\mathrm{a}_{11}
  • C 15 a1615 \,\mathrm{a}_{16}
  • D 14 a1614 \,\mathrm{a}_{16}

Step-by-Step Verified Solution

Let first term of G.P. be a and common ratio is r

Then, a1−r=5{a \over {1 - r}} = 5 ...... (i)

a(r5−1)(r−1)=9825⇒1−r5=98125a{{({r^5} - 1)} \over {(r - 1)}} = {{98} \over {25}} \Rightarrow 1 - {r^5} = {{98} \over {125}}

∴\therefore r5=27125, r=(35)35{r^5} = {{27} \over {125}},\,r = {\left( {{3 \over 5}} \right)^{{3 \over 5}}}

∴\therefore Then, S21=212[2×10ar+20×10ar2]{S_{21}} = {{21} \over 2}\left[ {2 \times 10ar + 20 \times 10a{r^2}} \right]

=21[10ar+10 . 10ar2]= 21\left[ {10ar + 10\,.\,10a{r^2}} \right]

=21 a11= 21\,{a_{11}}

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