Question

If nth term of an ap tn=(tn-1+3)then find few new terms of it.also find its sum of 66terms

Answer 1

Answer:

$n^{th}\text{ term of an AP}$ is given as

$t_{n}=t_{n-1}+3$

Let a be the first term of the AP and , d be the common difference of the A.P.

→a +(n-1) d= a + (n-1-1)d +3

→a + n d -d= a + n d - 2 d +3

→2 d -d =3

→d=3

Also, the difference between two consecutive terms gives common difference.

So, $t_{n}-t_{n-1}=3$

Which is common difference.

First term

$t_{1}=t_{1-1}+3\\\\ t_{1}=a=3$

$a_{1}=3,a_{2}=3+3=6,a_{3}=3+2*3=3+6=9,a_{4}=3+3*3=3+9=12$

$S_{n}=\frac{n}{2}[a+a_{n}]\\\\S_{66}=\frac{66}{2}*[3+3+(66-1)*3]\\\\S_{66}=33*[3+3-3+198]\\\\S_{66}=33*201=6633$