if the sum of first 7 terms of an ap is 49 and that of 17 terms is 289 find the sum of first n term
Answers
Answered by
2
Answer:
S=N/2(2a(n-1)d)
=n/2(-22+(n-1)×10)
n/2(-22+10n-10)
=n/2(-32 + 10n)
=-32n+10n^2/2
=2n(-16+5n)/2
=-16n+5n^2
wkt,
a+16d-a+6d= 289-49
10d=240
d=24
a=-11
Answered by
1
Answer:
Given
S
7
=49 and S
17
=289
By using S
n
=
2
n
[2a+(n−1)d] we have,
S
7
=
2
7
[2a+(7−1)d]=49
⇒49=
2
7
[2a+(7−1)d]
⇒49=
2
7
(2a+6d)
⇒7=a+3d
⇒a+3d=7...................(i)
S
17
=
2
17
[2a+(17−1)d]=289
⇒289=
2
17
[2a+(17−1)d]
⇒289=
2
17
(2a+16d)
⇒17=a+8d
⇒a+8d=17......................(ii)
Substituting (i) from (ii), we get
5d=10 or d=2
From equation (i),
a+3(2)=7
a+6=7 or a=1
S
n
=
2
n
[2(1)+(n−1)2]
=
2
n
[2+(n−1)2]
=
2
n
(2+2n−2)=n
2
Similar questions