Question

if the twice of the 2nd term is equal to 8 times of the 8 th term of an AP then the tenth term is ​

Answer 1

Answer:

5th term of an AP =t

5

=a+4d

1st term of an AP =t

1

=a

8th term of an AP =t

8

=a+7d

4th term of an AP =t

4

=a+3d

Given, t

5

=8t

1

⟹a+4d=8a

⟹4d=7a

⟹d=

4

7a

Also given, t

8

- 2t

4

=3

⟹a+7d−2(a+3d)=3

⟹d−a=3

Substituting d=

4

7a

in the above equation, we get a=4 and hence, d=7.

Answer 2

Answer:

10th term is '0'

Step-by-step explanation:

an A.P. is generally in the form of $a, a+d, a+2d, a+3d, .............$

or, let the 2nd term is $a+d$

and, thus the 8th term is $a+7d$

$or, 2(a+d)=8(a+7d)\\or, a+d=4(a+7d)\\or, a+d=4a+28d\\or, 4a-a=d-28d\\or, 3a=-27d\\or, a=-9d$         -----------(1)

or, the tenth term is, $a+(n-1)d= a+(10-1)d= a+9d$ = $-9d+9d=0$ (from $eqn.1.$ )