BITSAT 2022 Paper

Since, the first 11 terms are Ap,d=2
∴a11=a+10d=a+20
The middle term of AP is
T6=a+5d=a+10
For the next 11 terms in GP
r=2
∴ The middle term of GP is b(2)5 where, b is the first term of a GP which is the last term of AP
b(2)5=(a+20)32
According to the given condition,
⇒a+10=(a+20)32
⇒32a=10−640
a=−

630

31

∴ Middle term of entire sequence is 11th term
∴T11=

−630

31

+10×d
=

−630

31

+10×2=

−10

31