If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is:

 ❓ Question

If the sum of the 2nd, 4th and 6th terms of a G.P. of positive terms is 21 and the sum of its 8th, 10th and 12th terms is 15309, then find the sum of its first nine terms.

---

🖼️ Question Image

---

✍️ Short Solution

Let the G.P. be $a,\,ar,\,ar^{2},\dots$ with $a>0,\; r>0$
Write the given sums:

• Sum of 2nd, 4th, 6th terms:

  $$S_1=ar+a{r}^3+a{r}^5=ar(1+r^2+r^4)=21.$$

• Sum of 8th, 10th, 12th terms:

  $$S_2=a{r}^7+a{r}^9+a{r}^{11}=a{r}^7(1+r^2+r^4)=15309.$$

Divide $S_2$ by $S_1$ to eliminate the common factor $(1+r^2+r^4)\; and$$a r$:

$$\frac{S_2}{S_1}=\frac{a{r}^7(1+r^2+r^4)}{ar(1+r^2+r^4)}=r^6=\frac{15309}{21}\mathrm{.}$$

Compute the right-hand side:

$$\frac{15309}{21}=729.$$

So $r^{6} = 729 = 3^{6}$. Since $r>0$, we get

$$r=3.$$

Now substitute $r=3$ back into $S_1$:

$$ar(1+r^2+r^4)=ar(1+9+81)=ar\cdot 91=21.$$

Hence

$$ar=\frac{21}{91}=\frac{3}{13}\mathrm{.}$$

So

$$a=\frac{ar}{r}=\frac{3\mathrm{/}13}{3}=\frac{1}{13}\mathrm{.}$$

Now find the sum of first nine terms $S_9$. For $r\ne1$,

$$S_9=a\frac{r^9-1}{r-1}\mathrm{.}$$

We have $r^{9} = 3^{9} = 19683$, so

$$S_9=\frac{1}{13}\cdot \frac{19683-1}{3-1}=\frac{1}{13}\cdot \frac{19682}{2}=\frac{1}{13}\cdot 9841=\frac{9841}{13}\mathrm{.}$$

Divide: $9841 \div 13 = 757$. Therefore

$$\boxed{{\displaystyle S_9=757}}\mathrm{.}$$

---

🖼️ Image Solution

---

✅ Conclusion & Video Solution

Starting from the pair of 3-term sums we eliminated common factors to get $r=3$, found $a=\tfrac{1}{13}$, and then used the GP sum formula to compute the first nine-term sum.
Final answer:

$$\boxed{{\displaystyle \mathrm{757}\mathrm{<br>}\mathrm{<br>}}}$$