atulvaibhav1234
T6=0 a+5d=0 a=-5d----(i) now 33rd term=a+32d putting the value of a from equation (i) we get t33=-5d+32d =27d----(ii) now, t15=a+14d =-5d+14d =9d-----(iii) from equation (ii) and (iii) 3(t15)=(t33) proved. hope this helps you plz mark as brainliest.
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Here is ur answer
6th term = a + 5d = 0
a = -5d
33rd term = a + 32d
= (-5d) + 32d
= -5d + 32d
= 27d
15th term = a + 14d
= (-5d) + 14d
= -5d + 14 d
= 9d
33rd term = 27d
15th term = 9d
When compared,
27d = 3*9d
This means
33rd term = 3* 15th term
HENCE PROVED
a+5d=0
a=-5d----(i)
now
33rd term=a+32d
putting the value of a from equation (i) we get
t33=-5d+32d
=27d----(ii)
now,
t15=a+14d
=-5d+14d
=9d-----(iii)
from equation (ii) and (iii)
3(t15)=(t33)
proved.
hope this helps you
plz mark as brainliest.