Solution: Given that, The sum of G.P of 3 terms is 125 Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ $125=\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)...
Solution: Given that, The sum of GP $= 728$ Where, $r = 3, a = ?$ Firstly, $T_{n}=a r^{n-1}$ $486=a 3^{n-1}$ $486=a 3^{n} / 3$ $486(3)=a 3^{n}$ $1458=a 3^{n} \ldots .$ Eq. (i) $486=a 3^{n} / 3$...
Solution: Given that, The sum of GP $=381$ Where, $a=3, r=6 / 3=2, n=?$ Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ $\begin{array}{l} 381=3\left(2^{n}-1\right)...
Solution: Given that, The sum of GP $=39+13 \sqrt{3}$ Where, $a=\sqrt{3}, r=3 / \sqrt{3}=\sqrt{3}, n=?$ Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ $39+13...
Solution: Given that, The sum of GP $=728$ Where, $a=2, r=6 / 2=3, n=?$ Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ $\begin{array}{l} 728=2\left(3^{n}-1\right)...
Solution: Given that, The sum of G.P $=3069 / 512$ Where, $a=3, r=(3 / 2) / 3=1 / 2, n=?$ Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ $\begin{array}{l} 3069 /...
Solution: (i) $0.6+0.66+0.666+\ldots$ to $n$ terms. Let,s take 6 as a common term therefore we obtain, $6(0.1+0.11+0.111+\ldots n$ terms $)$ Now multiplying and dividing by 9 we obtain, $6 /...
Solution: (i) $9+99+999+\ldots$ to $n$ terms. We can write the given terms as $\begin{array}{l} (10-1)+(100-1)+(1000-1)+\ldots+n \text { terms } \\ \left(10+10^{2}+10^{3}+\ldots n \text { terms...
Solution: (i) $5+55+555+\ldots$ to $n$ terms. Let's take 5 as a common term therefore we obtain, $5[1+11+111+\ldots \mathrm{n}$ terms $]$ Now multiplying and dividing by 9 we obtain, $5 /...
Solution: (i) $\sum_{n=2}^{10} 4^{n}$ $=4^{2}+4^{3}+4^{4}+\ldots+4^{10}$ Where, $a=4^{2}=16, r=4^{3} / 4^{2}=4, n=9$ Using the formula, The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$...
Solution: (i) $\sum_{n=1}^{11}\left(2+3^{n}\right)$ $\begin{array}{l} =\left(2+3^{1}\right)+\left(2+3^{2}\right)+\left(2+3^{3}\right)+\ldots+\left(2+3^{11}\right) \\ =2 \times...
to 
terms;Solution: (i) $3 / 5+4 / 5^{2}+3 / 5^{3}+4 / 5^{4}+\ldots$ to $2 \mathrm{n}$ terms; We can write the series as: $3\left(1 / 5+1 / 5^{3}+1 / 5^{5}+\ldots\right.$ to $n$ terms $)+4\left(1 / 5^{2}+1 /...
to 5 terms; 
to 
termsSolution: (i) $2 / 9-1 / 3+1 / 2-3 / 4+\ldots$ to 5 terms; Given that $\begin{array}{l} a=2 / 9 \\ r=t_{2} / t_{1}=(-1 / 3) /(2 / 9)=-3 / 2 \\ n=5 \end{array}$ Using the formula, The sum of GP for...
to 8 terms; 
to 8 terms;Solution: (i) $0.15+0.015+0.0015+\ldots$ to 8 terms Given that $\begin{array}{l} a=0.15 \\ r=t_{2} / t_{1}=0.015 / 0.15=0.1=1 / 10 \\ n=8 \end{array}$ Using the formula, The sum of GP for $n$ terms...
to 10 termsSolution: (i) $4,2,1,1 / 2 \ldots$ to 10 terms It is known that, the sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ Given that, $\mathrm{a}=4, \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=2 /...
to Solution: (i) $1,-1 / 2,1 / 4,-1 / 8, \ldots$ It is known that, the sum of $\mathrm{GP}$ for infinity $=\mathrm{a} /(1-\mathrm{r})$ Given that, $\mathrm{a}=1, \mathrm{r}=\mathrm{t}_{2} /...
Solution: (i) $2,6,18, \ldots$ to 7 terms It is known that, the sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$ Given that, $a=2, r=t_{2} / t_{1}=6 / 2=3, n=7$ Substitute the values in...