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