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
. i.e.
T2+T6= ⇒ar+ar5= = . . . (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=()2 . . . (iii)
Divide Eq. (iii) by Eq. (ii),
= ⇒=⇒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
