The sum if the first seven terms and first seventeen terms of an AP are 49 and 286 respectively. find the sum of the first n terms of the AP. Please answer ASAP!
thanks in advance :)
suhanisharma:
Take your time guys , it took me quite a long time to get the answer and I still don't know if it's correct
um so sorry guys it's 289
Answers
Answered by
2
I think it is ncert question and it is 289 not 286
so here is the solution......
s7=49
s17= 289
7/2 [2a+(7-1)d]=49
2a+6d=14
a+3d=7.............(1)
17/2[2a+(17-1)d] =289
a+8d=17............(2)
subtracting (1) from (2)
we get...5d=10
d=2
From (1)....
a+3(2)=7
a=1
Sn = n/2[2a+(n-1)d]
=n/2[2*1+(n-1)*2]
=n/2(2+2n-2)
=n/2(2n)
=n^2
hope it helps you....
so here is the solution......
s7=49
s17= 289
7/2 [2a+(7-1)d]=49
2a+6d=14
a+3d=7.............(1)
17/2[2a+(17-1)d] =289
a+8d=17............(2)
subtracting (1) from (2)
we get...5d=10
d=2
From (1)....
a+3(2)=7
a=1
Sn = n/2[2a+(n-1)d]
=n/2[2*1+(n-1)*2]
=n/2(2+2n-2)
=n/2(2n)
=n^2
hope it helps you....
omg I'm so sorry ! u r right
I can't edit it right now
if it helps u please mark it as brainliest answer......
Yeah sure will
thank u so much
Answered by
2
let in an Ap , first term = a , common difference = d
Sn = n/2 [ 2a+(n-1)d]---(1)
1) given S7 = 49
⇒7/2[2a+6d]=49
⇒7[a+3d]=49
⇒a+3d=7-----(2)
2) S17 =289
⇒17/2[2a+16d]=289
⇒17[a+8d]=289
⇒a+8d=17----(3)
subtract (2) from (3)
5d=10
d=2
put d=2 in (2)
a+3*2=7
a=1
therefore put a=1 ,d=2 in (1)
Sn = n/2[2*1+(n-1)2]
=n[1+n-1]
=n*n
=n²
Sn = n/2 [ 2a+(n-1)d]---(1)
1) given S7 = 49
⇒7/2[2a+6d]=49
⇒7[a+3d]=49
⇒a+3d=7-----(2)
2) S17 =289
⇒17/2[2a+16d]=289
⇒17[a+8d]=289
⇒a+8d=17----(3)
subtract (2) from (3)
5d=10
d=2
put d=2 in (2)
a+3*2=7
a=1
therefore put a=1 ,d=2 in (1)
Sn = n/2[2*1+(n-1)2]
=n[1+n-1]
=n*n
=n²
Thanks a lot !! :)
:)
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