if seventh term of an AP is 1/9 and its ninth term is 1/7 find its 63rd term
Answers
Answered by
19
Hi ,
Let a and d are first term and
Common difference of an A.P
It is given that ,
7th term = a 7= 1/9
a + 6d = 1/9 ---( 1 )
a9 = 1/7
a + 8d = 1/7 ----( 2 )
Subtract ( 2 ) from ( 1 ) we get,
2d = 2/63
d = 1/63 ---( 3 )
Put d = 1/63 in equation ( 1 ) ,
we get
a + 6/63 = 1/9
a + 2/21 = 1/9
a = 1/9 - 2/21
a = ( 7 - 6 )/63
a = 1/63
Therefore ,
63 rd term = a63
a63 = a + 62d
= 1/63 + 62/63
= ( 1 + 62 )/63
= 63/63
a63 = 1
I hope this helps you.
: )
Let a and d are first term and
Common difference of an A.P
It is given that ,
7th term = a 7= 1/9
a + 6d = 1/9 ---( 1 )
a9 = 1/7
a + 8d = 1/7 ----( 2 )
Subtract ( 2 ) from ( 1 ) we get,
2d = 2/63
d = 1/63 ---( 3 )
Put d = 1/63 in equation ( 1 ) ,
we get
a + 6/63 = 1/9
a + 2/21 = 1/9
a = 1/9 - 2/21
a = ( 7 - 6 )/63
a = 1/63
Therefore ,
63 rd term = a63
a63 = a + 62d
= 1/63 + 62/63
= ( 1 + 62 )/63
= 63/63
a63 = 1
I hope this helps you.
: )
Answered by
5
◆◆◆◆
A7= 1/9
a + 6d = 1/9 ----(1)
A9 = 1/7
a + 8d = 1/ 7----(2)
subtract (1) from (2)
2d = 1/7 - 1/9
2d = 2 /63
d = 1/63
now put value of "d" in (1)
a + 6 × 1/63 = 1 /9
a = 1/ 9 - 6 / 63
a = 9 / 567
a = 1 / 63
now T63 = a + 62d
T63 = 1/ 63 + 62(1/63)
T63 = 63 /63 = 1
hence T63 = 1
◆hope it helps◆
A7= 1/9
a + 6d = 1/9 ----(1)
A9 = 1/7
a + 8d = 1/ 7----(2)
subtract (1) from (2)
2d = 1/7 - 1/9
2d = 2 /63
d = 1/63
now put value of "d" in (1)
a + 6 × 1/63 = 1 /9
a = 1/ 9 - 6 / 63
a = 9 / 567
a = 1 / 63
now T63 = a + 62d
T63 = 1/ 63 + 62(1/63)
T63 = 63 /63 = 1
hence T63 = 1
◆hope it helps◆
TheLostMonk:
what is the ans
one
haha ,then why you marked him brainliest ,
you marked him brainliest because he ranked genius and he cannot do mistakes,but ,
how to make ur solution brainliest
you can not make it correct , moderator calling me for correct my ans. but his ans is wrong
yes , i did a mistake.
thankyou ,you corrected it , moderator were calling for me to correct by ans, 2-3 times he called
Thank you .
ty ^_^
Similar questions