Question

गुणोत्तर श्रेणी 1+2/3+4/9+........के प्रथम n पदों का योग तथा प्रथम 5 पदों का योगफल ज्ञात कीजिए ?

Regards : Brainly Leader

Answer 1

Hay sir,

Given that ☞

$1 + \frac{2}{3} + \frac{4}{9} + ...$


Here, a =1 . r = 2/3.

गुणोत्तर श्रेणी के n पदों का योगफल,

$s _{n} = \frac{a(1 - r {}^{n} )}{1 - r}$



$= \frac{1(1 - ( \frac{2}{3} ) {}^{n}) }{1 - \frac{2}{3} } = 3(1 - ( \frac{2}{3} ) {}^{n} )$


और गुणोत्तर श्रेणी के प्रथम 5 पदों का योगफल


$s _{5} = 3(1 - ( \frac{2}{3} ) {}^{5} )$


$= 3(1 - \frac{32}{243} )$


$= 3 \times \frac{211}{243} = \frac{211}{81}$




I hope it's help you

Regards : Sangela

Answer 2

Given,
Series is :- 1+2/3+4/9+........

First term(a) = 1
common Ratio (r)= 2/3

We know that sum of n terms of GP is
Sn = a{1 - r^n}/(1-r)

putting a = 1 and. r= 2/3

Sn = 1*{1- (2/3)^n }/(1- 2/3}

Sn= 3*{1 - 2/3}^n}(3-2)

Sn = 3*{1 - 2/3}^n}.....................(i)
Now,
putting n=5 in equation (i), We get,
Sum of First 5 terms of GP = 3*{1 - (2/3)^5}

= 3{3^5 - 2^5}/3^5

= 1/3⁴{ 243 - 32}

=1/3⁴ {211}

= 211/3⁴

Note:-
(i) Sum of n term's of GP is = a{1 - r^n}/(1-r)
The following formula is used when r<1

(ii) Sum of n terms of GP is = a{r^n -1 }/(r- 1)
The following formula is used when r>1