Question

7th term of an ap is -4 and 30th term is -16 Ap is

Answer 1

⭐Given:

• t7= -4

• t30= -16

⭐To Find:

• Arithmetic Progression=?

⭐Solution:

Since,

We know that,

tn=a+(n-1) d......... (formula)

: . t7= a+(7-1) d

: . a+6d=-4............ (1)

And,

t30=a+(30-1) d

: . a+29d=-16............ (2)

Subtracting equation (1) from (2) we get,

23d =-12d.

: . d= -23/12

Now, Substituting the value of d in (1) we get,

a+6(-23/12) =-4

: . a-23/2= -4

: . a= -4+23/2

: . a= -8+23/2

: . a= 15/2

Now Here a=t1=15/2

t2=a+d=15/2+(-23/12)

= 15/2-23/12

=90/12-23/12

= 90-23/12

= 67/12

t3=t2+d=67/12+(-23/12)

=67/12-23/12

=67-23/12

=44/12

=11/3

Therefore, The A. P is 15/2, 67/12, 11/3.

Answer 2

Answer:

ANSWER

3, 15, 27, 39…….

a

n

=a+(n−1)d

a=3,d=15−3=12

n=54

a

54

=a+(n−1)d

=3+(54−1)×12

=3+53×12

=639

Term which is 132 more than its 54th term is – 639 +132 = 771.

a

n

=771

771=a+(n−1)d

771=3+(n−1)12

771=3+(n−1)12

64=n−1

n=65