Let the sum of n, 2n, 3n terms of an A.P. be S1, S2, and S3, respectively, show that
S3 = 3(S2 – S1)
Answers
Let a and b be the first term and the common difference of the A.P. respectively.
Therefore,
S1 = (n/2)[2a + (n - 1)d] …(1)
S2 = (2n/2)[2a + (2n - 1)d] …(2)
S3 = (3n/2)[2a + (3n - 1)d] …(3)
From (1) and (2), we obtain
S2 - S1 = (2n/2)[2a + (2n - 1)d] - (n/2)[2a + (n - 1)d]
= (n/2)[{4a + (4n - 2)d} - {2a + (n - 1)d}]
= (n/2)[4a + 4nd - 2d - 2a - nd + d]
= (n/2)[2a + 3nd - d]
= (1/3) × (3n/2)[2a + (3n - 1)d]
= (1/3)S3
Thus, S3 = 3(S2 - S1)
Hence, the given result is proved.
JAI SHREE KRISHNA
Answer:
Let a and d be the first term and the common difference of the A.P. respectively.
Therefore,
S
1
=
2
n
[2a+(n−1)d]....(1)
S
2
=
2
2n
[2a+(2n−1)d]=n[2a+(n−1)d]...(2)
S
3
=
2
3n
[2a+(3n−1)d]...(3)
From (1) and (2), we obtain
S
2
−S
1
=n[2a+(2n−1)d]−
2
n
[2a+(n−1)d]
=n{
2
4a+4nd−2d−2a−nd+d
}
=n[
2
2a+3nd−d
]
=
2
n
[2a+(3n−1)d]
∴3(S
2
−S
1
)=
2
3n
[2a+(3n−1)d]=S
3
[From(3)] [henceproved]