Question
For a given $G.P.$, if $T_2=9$ and $T_5=243$; then find $S_4$.
✓
Answer
Here, $T_2=9 ; T_5=243 ; S_4=$ ?
$T_n=a . r^{n-1}$
$\therefore T_2=a . r^{2-1}$
$\therefore 9=a r..........(1)$
$T_5=a . r^{5-1}$
$\therefore 243=a r^4.........(2)$
Taking ratio of results $(2)$ and $(1)$,
$\frac{243}{9}=\frac{a r^4}{ a r }$
$\therefore 27=r^3$
$\therefore(3)^3=r^3$
$\therefore r=3$
Putting $r=3$ in the result $(1),$
$9=3 a \therefore a=3$
Now, $S_n=\frac{a\left[r^n-1\right]}{r-1}(\because r > 1)$
$\therefore S_4=\frac{3\left[(3)^4-1\right]}{3-1}=\frac{3\lfloor 81-1]}{2}=\frac{3 \times 80}{2}=120$
$T_n=a . r^{n-1}$
$\therefore T_2=a . r^{2-1}$
$\therefore 9=a r..........(1)$
$T_5=a . r^{5-1}$
$\therefore 243=a r^4.........(2)$
Taking ratio of results $(2)$ and $(1)$,
$\frac{243}{9}=\frac{a r^4}{ a r }$
$\therefore 27=r^3$
$\therefore(3)^3=r^3$
$\therefore r=3$
Putting $r=3$ in the result $(1),$
$9=3 a \therefore a=3$
Now, $S_n=\frac{a\left[r^n-1\right]}{r-1}(\because r > 1)$
$\therefore S_4=\frac{3\left[(3)^4-1\right]}{3-1}=\frac{3\lfloor 81-1]}{2}=\frac{3 \times 80}{2}=120$
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