if the 5th term of ap is zero,show that 33rd term is 4 times 12th TERM?
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From the the normal formula of Arithmetic Progression,
T=a+(n-1)d
now for the 5th term(n=5)
T5=a+4d
0=a+4d
d=(-a/4)............. (equation first)
now for the 12th term(n=12)
T12=a+11d
from equation first,
T12=a+11(-a/4)
T12=(4a-11a)/4= (-7a/4)............ (equation second)
now for the 33rd term(n=33)
T33=a+32d
from equation first
T33=a+32(-a/4)
T33=(4a-32a)/4
T33=(-28a/4)
now comparing with equation second,
T33=4(T12)
T=a+(n-1)d
now for the 5th term(n=5)
T5=a+4d
0=a+4d
d=(-a/4)............. (equation first)
now for the 12th term(n=12)
T12=a+11d
from equation first,
T12=a+11(-a/4)
T12=(4a-11a)/4= (-7a/4)............ (equation second)
now for the 33rd term(n=33)
T33=a+32d
from equation first
T33=a+32(-a/4)
T33=(4a-32a)/4
T33=(-28a/4)
now comparing with equation second,
T33=4(T12)
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