Question

2. ni Given that the nth term, T., of a sequence is , T, = 2n +5, find n (i) the 5th term (ii) the 8th term (iii) the lowest common multiple of the 5th and the 8th term, of the sequence. 1

​

Answer 1

Answer:

Tn=Sn−Sn−1      .(1)

Now Sn=3n2+4n

∴Sn−1=3(n−1)2+4(n−1)=3n2−2n−1

∴Tn=(3n2−4n)−(3n2−2n−1), by (1)

=6n+1

∴T1=7,T2=13,T3=19,______

∴a=7,d=6.

Answer 2

Given n th term=2n+5

1) 5th term
2(5)+5=15

2) 8th term
2(8)+5=21

3)LCM of 15,21
7x3x5=105