Math, asked by kailashambedkar, 1 year ago

Please solve the 20th sum

Attachments:

Answers

Answered by zahidpatel

For 1st AP
a=8
d=20
Sn1=n/2 {2a+(n-1)d}
=n/2{2×8+(n-1)20
=n/2{16+20n-20
=n/2{20n-4} ........eq 1

For 2nd AP
a=-30
d=8
Sn2=n/2{2a+(n-1)d}
=n/2{-60+8n-8}
=n/2{8n-68} ..........eq 2

given
Sn1=Sn2
n/2{20n-4}=n/2{8n-68}
20n-4=8n-68
20n-8n=68+4
12n=72
n=6

hope it helps u

isha7928: good night

isha7928: please help in my chemistry question which i asked in morning

zahidpatel: oh so sad for u

zahidpatel: i will try today to prupose her

zahidpatel: means alaina

zahidpatel: plzz mark my answer as brainlist of tangent secant theorem

isha7928: all the best

zahidpatel: thanx

zahidpatel: mar as brain list my answer

isha7928: kia hai

Answered by samiksha2259

Answer:

Step-by-step explanation:

Sn=S2n........... (given)

By the solving sn term.....

We get....

n/2(20n-4)....... equation no 1

By the solving s2n term.....

We get....

n/2(8n-68)....... equation no 2

n/2(20n-4) =n/2(8n-68)...... (according the given condition)

Then we get......

20n-4=8n-68

20n-8n=-68+4

-12n=-64

samiksha2259: Sorry..... I am writing and unwantedly it send.....

samiksha2259: n=64/12

samiksha2259: n=5.333333333

Previous Question

Next Question