Question

find the 8th term of sequence whose nth term is (n+8) /(n+4) ×[1+3^n /2^n+3^n]​

Answer 1

Answer:

Given 3,3,6………

Now 3,3,6,9,15,24,39,63,102

(It involves smaller numbers)

3

3+0=3

3+3=6

3+6=9

6+9=15

9+15=24

15+24=39

24+39=63 (8

th

term)

∴63 is 8

th

term.

Answer 2

Answer:

Given, an=3+32n

Now, an+1=3+32n+1

∴ common difference d=an+1−an

                                         =3+32n+1−3−32n

                                         =32

First term, a=a1=3+32

                     =311

Sum of n term of an arithmetic progression is 

    Sn=2n[2a+(n−1)d]

⇒S24=224[322+23(32)]

           =12[322+46]

           =312×68

           =4×68

           =272