Verified Solution • 2025–26 Session moderate

QJEEM21Ma5724: JEE Main 2021 Mathematics - Class 11 JEEM21Ma5724 Solution: Let a 1 , a 2 , a 3 , ..... be an A.P. If , p 10, then is equal …

Let a1, a2, a3, ..... be an A.P. If a1+a2+....+a10a1+a2+....+ap=100p2{{{a_1} + {a_2} + .... + {a_{10}}} \over {{a_1} + {a_2} + .... + {a_p}}} = {{100} \over {{p^2}}} , p ≠\ne 10, then a11a10{{{a_{11}}} \over {{a_{10}}}} is equal to :

Multiple Choice Options

  • A 1921{{19} \over {21}}
  • B 100121{{100} \over {121}}
  • C 2119{{21} \over {19}}
  • D 121100{{121} \over {100}}

Step-by-Step Verified Solution

102(2a1+9d)p2(2a1+(p−1)d)=100p2{{{{10} \over 2}(2{a_1} + 9d)} \over {{p \over 2}(2{a_1} + (p - 1)d)}} = {{100} \over {{p^2}}}

(2a1+9d)p=10(2a1+(p−1)d)(2{a_1} + 9d)p = 10(2{a_1} + (p - 1)d)

9dp=20a1−2pa1+10d(p−1)9dp = 20{a_1} - 2p{a_1} + 10d(p - 1)

9p=(20−2p)a1d+10(p−1)9p = (20 - 2p){{{a_1}} \over d} + 10(p - 1)

a1d=(10−p)2(10−p)=12{{{a_1}} \over d} = {{(10 - p)} \over {2(10 - p)}} = {1 \over 2}

∴\therefore a11a10=a1+10da1+9d=12+1012+9=2119{{{a_{11}}} \over {{a_{10}}}} = {{{a_1} + 10d} \over {{a_1} + 9d}} = {{{1 \over 2} + 10} \over {{1 \over 2} + 9}} = {{21} \over {19}}

Related Learning Resources & Exam Preparation