3 Marks Question

Question 513 Marks

How many terms of G.P. $3, 3^2 , 3^3 ......$ are needed to give the sum $120?$

Answer

Here, $a = 3$ and r = $ \frac { 3 ^ { 2 } } { 3 } = 3$
$\therefore ^ { \mathrm { S } _ { n } = \frac { a \left( r ^ { - } - 1 \right) } { r - 1 } }$ when $r > 1$
$\Rightarrow 120 = \frac { 3 \left( 3 ^ { n } - 1 \right) } { 3 - 1 }$
$\Rightarrow 120 = \frac { 3 } { 2 } \left( 3 ^ { n } - 1 \right)$
$ \Rightarrow 120 \times \frac { 2 } { 3 } = 3 ^ { n } - 1$
$ \Rightarrow 3^n = 81$
$\Rightarrow 3^n = (3)^4$
$\Rightarrow n = 4$
Therefore, the sum of $4$ terms of the given G.P. is $120$.

View full question & answer→

Question 523 Marks

The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is $1$. Find the common ratio and the terms.

Answer

Let $\frac ar , a, r$ be first three terms of the given G.P.According to question$,\frac { a } { l ^ { r } } + a + a r = \frac { 39 } { 10 }……….(i)$
And$\frac { a } { r } \times a \times r = 1$
$\Rightarrow a ^ { 3 } = 1$
$\Rightarrow a = 1$
$Putting value of a in eq. (i),$
$\Rightarrow 10+ 10r + 10r^2 = 39r$
$\Rightarrow 10r^2- 29r +10 = 0$
$\Rightarrow r = \frac { - ( - 29 ) \pm \sqrt { ( - 29 ) ^ { 2 } - 4 \times 10 \times 10 } } { 2 \times 10 }$
$\Rightarrow r = \frac { 29 \pm \sqrt { 841 - 400 } } { 20 }$
$\Rightarrow r = \frac { 29 \pm 21 } { 20 }$
Taking $r = \frac { 29 + 21 } { 20 } = \frac { 50 } { 120 } = \frac { 5 } { 2 }$ and
then the first three terms are $\frac { 1 } { 5 / 2 } , 1,1 \times \frac { 5 } { 2 }$
$\Rightarrow \frac { 2 } { 5 } , 1 , \frac { 5 } { 2 }$
Taking $r = \frac { 29 - 21 } { 20 } = \frac { 8 } { 20 } = \frac { 2 } { 5 }$
then first three terms are $\frac { 1 } { 2 / 5 } , 1,1 \times \frac { 2 } { 5 }$
$\Rightarrow\frac { 5 } { 2 } , 1 , \frac { 2 } { 5 }$

View full question & answer→

Question 533 Marks

Evaluate:$\sum \limits_ { k = 1 } ^ { 11 } \left( 2 + 3 ^ { k } \right)$

Answer

Given:$\sum _ { k = 1 } ^ { 11 } \left( 2 + 3 ^ { k } \right)$
$= (2 + 3^1) + (2 + 3^2) + (2 + 3^3) + (2 + 3^{11})$
$= ( 2 + 2 + 2 +........11\ \text{times}) + (3 + 3^2 + 3^3 +....... +3^{11})= 22 + (3 + 3^2 + 3^3 +....... +3^{11}) ……….(i)$
Here $3, 3^2,3^3 ....... ,3^{11}$ is in G.P.
$\therefore$a = 3 and r = $\frac { 3 ^ { 2 } } { 3 } = 3$
$\mathrm { S } _ { n } = \frac { 3 \left( 3 ^ { 11 } - 1 \right) } { 3 - 1 } = \frac { 3 } { 2 } \left( 3 ^ { 11 } - 1 \right)$
Putting the value of $S_n$ in eq. (i), we get $\sum _ { k = 1 } ^ { 11 } \left( 2 + 3 ^ { k } \right) = 22 + \frac { 3 } { 2 } \left( 3 ^ { 11 } - 1 \right)$

View full question & answer→

Question 543 Marks

Find the sum to indicated number of terms of the geometric progression $x^3, x^5, x^7 ... n$ terms $($if $x \ne \pm1)$.

Answer

Here,$a = x^3$ and $r = \frac { x ^ { 5 } } { x ^ { 3 } } = x^2$
$S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }$ when $r < 1$
$\Rightarrow S _ { n } = \frac { x ^ { 3 } \left[ 1 - \left( x ^ { 2 } \right) ^ { n } \right] } { 1 - x ^ { 2 } }$
$\Rightarrow \mathrm { S } _ { n } = \frac { x ^ { 3 } } { 1 - x ^ { 2 } } \left[ 1 - x ^ { 2 n } \right]$

View full question & answer→

MATHS 3 Marks Question