JEE Main Math Class 11 Sequences and Series Part 2 Questions

Let the first term of geometric series be' a′ and common ratio be 'r'.
Then, n th term of given series is given as
Tn=arn−1
Now, given that sum of second and sixth term is 25/2.
i.e. T2+T6=25∕2
⇒ar+ar5=25∕2
⇒ar(1+r4)=25∕2‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Also, given that product of third and fifth term is 25.
lrl‌ i.e. ‌‌(T3)(T5)=25
⇒‌(ar2)(ar4)=25
⇒‌a2r6=25‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
Squaring Eq. (i), we geta2r2(1+r4)2=(‌

25

2

)2‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)
Divide Eq. (ii) by (iii),
⇒‌‌‌

a2r2(1+r4)2

a2r6

=‌

(25)2

4(25)

⇒‌‌‌

(1+r4)2

r4

=‌

25

4

⇒‌‌4(1+r4)2=25r4
⇒‌‌4(1+r8+2r4)=25r4
⇒‌‌4r8−17r4+4=0
⇒‌‌4r8−16r4−r4+4=0
⇒‌‌4r4(r4−4)−1(r4+(−4))=0
⇒‌‌‌‌(r4−4)(4r4−1)=0
‌ Gives, ‌r4=40rr4=1∕4
We have to find sum of 4 th, 6th and 8th term, i.e.
T4+T6+T8‌‌=ar3+ar5+ar7
‌‌=ar(r2+r4+r6)
‌‌=ar3(1+r2+r4)‌‌‌⋅⋅⋅⋅⋅⋅⋅(iv)
Using Eq. (ii),
(ar3)2=25
⇒ar3=5
Also, we take r4=4 because given series is increasing and r2=2 .
∴‌‌T4+T6+T8‌‌=5(1+2+4)
‌‌=5(7)=35