Question

If S50=S30 in a AP find S50=?

Answer 1

Answer:

$S_{80}=0$

Step-by-step explanation:

Formula used:

The sum of first n terms of an A.P is

$\boxed{\bf\:S_n=\frac{n}{2}[2a+(n-1)d]}$

Given:

$S_{50}=S_{30}$

$\implies\frac{50}{2}[2a+(50-1)d]=\frac{30}{2}[2a+(30-1)d]$

$\implies\:25[2a+49d]=15[2a+29d]$

$\implies\:5[2a+49d]=3[2a+29d]$

$\implies\:10a+245d=6a+87d$

$\implies\:4a+158d=0$

$\implies\:2a+79d=0$

$\implies\:40[2a+79d]=0$

$\implies\:\frac{80}{2}[2a+(80-1)d]=0$

$\implies\:\bf\:S_{80}=0$

Answer 2

$S_{80}=0$

Step-by-step explanation:

Given: In A.P

$S_{50}=S_{30}$

Sum of nth terms of an A.P

$S_n=\frac{n}{2}(2a+(n-1)d)$

Where n=Total number of terms

a=First term of  A.P

d=Common difference

Using the formula

$\frac{50}{2}(2a+49d)=\frac{30}{2}(2a+29d)$

$25(2a+49d)=15(2a+29d)$

$2a+49d=\frac{15}{25}(2a+29d)$

$2a+49d=\frac{3}{5}(2a+29d)$

$10a+245d=6a+87d$

$10a-6a+245d-87d=0$

$4a+158d=0$

$2(2a+79d)=0$

$2a+79d=0$

$S_{80}=\frac{80}{2}(2a+79d)$

$S_{80}=40\times 0=0$

#Learn more:

https://brainly.in/question/75002:Answered by Pranavchoudhary