Answer:

t

3

​

=34 and t

8

​

=69

(a) Now, t

n

​

=a+(n−1)d

⇒t

3

​

=a+(2−1)d and t

8

​

=a+(7−1)d

⇒34=a+(2−1)d and 69=a+(7−1)d

⇒34=a+d and 69=a+6d

Thus, 69−34=a+6d−a−d

⇒35=5d⇒d=7

(b) First term, a=t

3

​

−2d=34−2(7)=34−14=20

General term, t

n

​

=a+(n−1)d=20+(n−1)7=20+7n−7=7n+13

Thus, t

n

​

=7n+13 is the algebraic form of this sequence.

(c) t

n

​

=7n+13

If each term of the sequence is multiplied by 4 and then 3 is added to it, the

general term will be t

n

​

=[(7n+13)×4]+3=28n+52+3=28n+55

Thus, tenth term, t

10

​

=28(10)+55=280+55=335

Step-by-step explanation:

Verified answer

Answer:

a3= 13

[tex] \red{ = > a + 2d = 13...(1)}[/tex]

a6= 28

[tex] \orange {= > a + 5d = 28..(2)}[/tex]

Subtract these equations:-

[tex]3d = 15 \\ \pink{ \:d = 5}[/tex]

Put in eq. 1

[tex] \blue{a = 3}[/tex]