If first term is 3 and common ratio is 3 then find the 6th term of G.P

Explanation

Let `P(n):(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+.....+(1)/((3n-1)(3n+2))=(n)/(6n+4)` ......(i)
Step I For `n=1`,
LHS of Eq. (i) `=(1)/(2.5)=(1)/(10)`
RHS of Eq. (i) `=(1)/(6xx1+4)=(1)/(10)`
LHS = RHS
Therefore , P(1) is true .
Step II Let us assume that the result is true for `n=k`. Then , `P(k):(1)/(2.5)+(1)/(5.8)+(1)/(11)+.....+(1)/((3k-1)(3k+2))+(1)/((3k+2)(3k+5))`
`=((k+1))/(6(k+1)+4)=((k+1))/(6k+10)`
LHS `=(1)/(2.5)+(1)/(5.8)+(1)/(8.11)+....+(1)/((3k-1)(3k+2))+(1)/((3k+2)(3k+5))`
`=(k)/(6k+4)+(1)/((3k+2)(3k+5))` [by assumption step]
`=(k(3k+5)+2)/(2(3k+2)(3k+5))=(3k^2+5k+2)/(2(3k+2)(3k+5))`
`=((k+1)(3k+2))/(2(3k+2)(3k+5))=(k+1)/(6k+10)=RHS`
This shows that the result is true for `n=k+1`. Therefore , by the principle of mathematical induction , the result is true for all ` n in N`.