If the sum of first 7 terms of an Arithmetic progression is 49 and that of first 17 terms of 289, then find the sum of the first 'n' terms .
Answers
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
Answer:
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 terms. Here S7 = 49, S17 = 289 Let 'a' be the first term and 'd' the common difference. The sum of the third and the seventh terms of an AP is 6 and their product is 8.
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