Question

in a certain AP , 2 times the 4th term is equal to 3 times with the 7th term , find the value of the 13th term

Answer 1

Answer:

a13 = 0 If u got Your answer please like

Answer 2

Answer:

The nth term of an A.P is given by

a n =a1+(n−1)d

where d is the common difference

and  a 1​is the first term of A.P

hence 4th term of an A.P is a 4

​=a 1+(4−1)d⟹a 4

=a1 +3d….eq(1)

given that 4th term of A.P is three times the first .

⟹a4=3a 1

put value of a 4

 from eq(1)

⟹a 1+3d=3a1

⟹3a 1−a 1

​=3d⟹2a 1​

=3d⟹a 1

=23d

….eq(2)

3rd term of A.P is given by

a =a 1+(3−1)da 3

=a 1+2d

7th term of A.P is given by

a 7 =a 1+(7−1)d

a 7=a 1 +6d

given that 7th term exceeds twice the third by 1

⟹a 7

=2a 3+1

put values of a 7

​and a 3a 1+6d=2(a 1 +2d)+1

⟹2a 1

​−a 1

=6d−4d−1

⟹a 1

=2d−1…..eq(3)

put value of a 1from eq(2) to eq(3)

⟹23d

=2d−1

⟹3d=4d−2

⟹4d−3d=2

⟹d=2...…..eq(4)  common difference of A.P

put value of d from eq(4) to eq(2) we get

⟹a 1​ = 2

3​⟹a 1

=3......eq(5)  first term of A.PThe nth term of an A.P is given by

a n =a 1+(n−1)d

where d is the common difference

and  a 1

is the first term of A.P

hence 4th term of an A.P is a 4​

=a 1 +(4−1)d

⟹a 4=a 1+3d….eq(1)

given that 4th term of A.P is three times the first .

⟹a 4=3a 1

put value of a

4from eq(1)

⟹a 1 +3d=3a 1

⟹3a 1 −a 1=3d

⟹2a 1 =3d

⟹a1= 23d

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