- Home
- Standard 11
- Mathematics
8. Sequences and Series
hard
Consider an $A.P.$ of positive integers, whose sum of the first three terms is $54$ and the sum of the first twenty terms lies between $1600$ and $1800$ . Then its $11^{\text {th }}$ term is :
A$84$
B$122$
C$90$
D$108$
(JEE MAIN-2025)
Solution
$S_3=3 a+3 d=54$
$\Rightarrow a+d=18$
$S_{20}=10(2 a+19 d)$
$\Rightarrow 10(36+17 d)$
$\Rightarrow 1600 < 10(36+17 d) < 1800$
$\Rightarrow 160 < 36+17 d < 180$
$\Rightarrow 124 < 17 d < 144$
$\Rightarrow 7 \frac{5}{17} < d < 8 \frac{8}{17}$
Common difference will be natural number
$\Rightarrow d=8 \Rightarrow a=10$
$\Rightarrow a_{11}=10+10 \times 8=90$
$\Rightarrow a+d=18$
$S_{20}=10(2 a+19 d)$
$\Rightarrow 10(36+17 d)$
$\Rightarrow 1600 < 10(36+17 d) < 1800$
$\Rightarrow 160 < 36+17 d < 180$
$\Rightarrow 124 < 17 d < 144$
$\Rightarrow 7 \frac{5}{17} < d < 8 \frac{8}{17}$
Common difference will be natural number
$\Rightarrow d=8 \Rightarrow a=10$
$\Rightarrow a_{11}=10+10 \times 8=90$
Standard 11
Mathematics