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Question: The first term of an infinite G.P. is \(1\) and any term is equal to the sum of all the succeeding t...

The first term of an infinite G.P. is 11 and any term is equal to the sum of all the succeeding terms. Find the sum of the infinite series.

Explanation

Solution

Hint: Any term in G.P. is equal to the sum of all the succeeding terms. We have:
Tn=Tn+1+Tn+2+Tn+3+......\Rightarrow {T_n} = {T_{n + 1}} + {T_{n + 2}} + {T_{n + 3}} + ......\infty

The general term of G.P. can be written as:
Tn=arn1.....(i)\Rightarrow {T_n} = a{r^{n - 1}} .....(i)
And according to the information given in the question, any term of the G.P. is equal to the sum of all the succeeding terms. From this we’ll get:
Tn=Tn+1+Tn+2+Tn+3+......\Rightarrow {T_n} = {T_{n + 1}} + {T_{n + 2}} + {T_{n + 3}} + ......\infty
Substituting corresponding values in equation(i)(i), we’ll get:
arn1=arn+arn+1+arn+2+......,\Rightarrow a{r^{n - 1}} = a{r^n} + a{r^{n + 1}} + a{r^{n + 2}} + ......\infty ,
aa is the first term of G.P. and its value is 11 as per the information given in the question. So putting its value, we’ll get:
rn1=rn+rn+1+rn+2+....., rn1=rn[1+r+r2+.....], 1r=[1+r+r2+.....].....(ii)  \Rightarrow {r^{n - 1}} = {r^n} + {r^{n + 1}} + {r^{n + 2}} + .....\infty , \\\ \Rightarrow {r^{n - 1}} = {r^n}\left[ {1 + r + {r^2} + .....\infty } \right], \\\ \Rightarrow \dfrac{1}{r} = \left[ {1 + r + {r^2} + .....\infty } \right] .....(ii) \\\
Now, the terms on the right hand side of the above equation constitutes an infinite G.P. with 11 as the first term and rr as the common ratio. And we know the formula for sum of terms of infinite G.P.:
S=a1r\Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}
So, on using this formula for equation (ii)(ii),we’ll get:
1r=11r, 1r=r, 2r=1, r=12.  \Rightarrow \dfrac{1}{r} = \dfrac{1}{{1 - r}}, \\\ \Rightarrow 1 - r = r, \\\ \Rightarrow 2r = 1, \\\ \Rightarrow r = \dfrac{1}{2}. \\\
Thus, the common ratio of the G.P. is 12\dfrac{1}{2} and its first term is already given as 11. So, we our infinite G.P.:
1,12,14,18,.......\Rightarrow 1,\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},.......\infty
For finding sum of its terms, we will again applyS=a1r{S_\infty } = \dfrac{a}{{1 - r}}, we’ll get:
S=1112, S=2  \Rightarrow {S_\infty } = \dfrac{1}{{1 - \dfrac{1}{2}}}, \\\ \Rightarrow {S_\infty } = 2 \\\
Therefore, the sum of infinite G.P. is 22.

Note: If a G.P. consists of infinite terms, then we can only calculate the sum of its terms if it's common ratio is greater than 00 and less than 11 (0<r<1)\left( {0 < r < 1} \right).Otherwise its sum will not be defined.