Question

If the 10th term of an AP is 21 and the sum of its first ten terms is 120, find its nth term....


Please guys give me answer quickly.. ​

Answer 1

Answer:

your answer is here buddy....

given,

a10= 21

s10= 120

to find ,

an.....

a10= a+(n-1)d

21= a+9d

a+9d-21= 0.............(1)

s10= n/2(2a+(n-1)d

120= 5(2a+9d)

120= 10a+45d

10a+45d-120= 0..........(2)

solve 1 & 2...by elimination method...

10a+90d=210...........(multiply by 10)

10a+45d= 120.......

now, change the sign..

10a+90d= 210

-10a -45d= -120

__. 45d= 90

d= 90/45

d=2

then a,

a+9d= 21

a+18= 21

a= 3

now applying an..formulla,

an = a+(n-1)d

taking n as 2.......(let it be 2nd term)

a2= 3+1(2)

a2= 5......

then an =....

an= 3+(n-1)2

an= 3+2n-2

an= 1+2n”...

hope this will help you...

plss mark as brainlist....

Answer 2

Answer:

a10=21

s10=120

1. an=a+(n-1)d=?
2. a10=a+95=21

• s10=n÷2{2a (n-1)d)]=120
• 10÷2 [2a+(10-1)d]=120
• 5 [2a+9d ]=120
• 2a+9d=120÷5
• 2a+9d=24
• from 1 and 2 equation sub:

1. a+9d=21
2. a+9d=24

• - - -
• -a= -3
• a=3

1. a+9d=21

• 3+9d=21
• 9d=21-3
• 9d=18
• d=18÷9
• d=21

an=a+(n-1)d

= 3+(n-1)2

= 3+2n-2

= 2n+1

hence proof