Question

If the 5th yerm of an AP is zero , show that its 33rd term is four times its 12th term

Answer 1

Hii There!!

Since, a5= a+4d=0----------1) [ since,an=a+(n-1)d]

=) a+32d = 4[a+11d]. [ {a+(33-1)d} and {a+(12-1)d } ]

=) a+32d= 4a+44d

=) 3a+12d =0

=) a +4d=0 ---------2)

Hence, eq.1) and 2) are equal ..

Hence proved..


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Answer 2

Hii friend,

Let the first term of AP be a.

And,

common difference be D.

Tn = a+(n-1) × d

T5 = a + (5-1) × d

=> a +5d-d = a+4d



T33 = a +(33-1) × d

T33 => a+33d-d = a+32d

And,


T12 = a+(12-1) × d

T12 = a+12d-d = a+11d

Now,

T5 = 0

a+4d = 0

a = -4d

Therefore,

T33 = a+32d = -4d +32d = 28d

and,

T12=> a+11d = -4d +11d = 7d

According to question,

28d = 4(7d)

28d = 28d.....PROVED.....

Hence,

33 term of the is equal to 4 times it's 12th term


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