MCQ
The difference between the fourth term and the first term of a Geometrical Progresssion is $52.$ If the sum of its first three terms is $26,$ then the sum of the first six terms of the progression is
- A$63$
- B$189$
- ✓$728$
- D$364$
According to given condition, we have
$a{r^3} - a = 5 \Rightarrow a\left( {{r^3} - 1} \right) = 52\,\,\,\,......\left( 1 \right)$
and $a + ar + a{r^2} = 26$
$ \Rightarrow a\left( {1 + r + {r^2}} \right) = 26\,\,\,\,\,\,......\left( 2 \right)$
To find : $a\left( {1 + r + {r^2} + {r^3} + {r^4} + {r^5}} \right)$
Consider
$a\left[ {1 + r + {r^2} + {r^3} + {r^4} + {r^5}} \right]$
$ = a\left[ {1 + r + {r^2} + {r^3}\left( {1 + r + {r^2}} \right)} \right]$
$ = a\left[ {1 + r + {r^2}} \right]\left[ {1 + {r^3}} \right]\,\,\,\,\,\,......\left( 3 \right)$
Divide $(1)$ by $(2)$, we get
$\frac{{{r^3} - 1}}{{1 + r + {r^2}}} = 2$
we know ${r^3} - 1 = \left( {1 + {r^3}} \right)\left( {1 + r + {r^2}} \right)$
$\therefore r - 1 = 2 \Rightarrow r = 3\,\,\,\,$ and $a = 2$
$\therefore a\left( {1 + r + {r^2} + {r^3} + {r^4} + {r^5}} \right)$
$ = a\left( {1 + r + {r^2}} \right)\left( {1 + {r^3}} \right)$
$ = 2\left( {1 + 3 + 9} \right)\left( {1 + 27} \right)$
$ = 26 \times 28 = 728$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ $S Q_1=2$
$(B)$ $Q _1 Q _2=\frac{3 \sqrt{10}}{5}$
$(C)$ $PQ _1=3$
$(D)$ $SQ _2=1$