Sequences and Series Part 2

Let the first term of geometric series be ' a ' and common ratio be ' r′.
Then, nth 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

=‌

25

2

. . . (i)
Also, given that product of third and fifth term is 25 . i.e. ‌‌(T3)(T5)=25
⇒‌‌(ar2)(ar4)=25
⇒ a2r6=25 . . . (ii)
Squaring Eq. (i), we get
a2r2(1+r4)2=(‌

25

2

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

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
We have to find sum of 4 th, 6 th and 8 th term, i.e.
T4+T6+T8‌‌=ar3+ar5+ar7
‌‌=ar(r2+r4+r6)
‌‌=ar3(1+r2+r4). . . (iv)
Using Eq. (ii),
‌ t ‌(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