Question

If the sum of first n, 2n, 3n term of an ap be s1, s2, s3 respectively then prove that s3 = 3(s2 – s1).

Answer 1

S2=2n/2[2a+(2n-1)d] ..(I)

S1=n/2[2a+(n-1)d]..(ii)

S3=3n/2[2a+(3n-1)d]...(iii)

Subtracting eq. (ii) from (I) we get

=>S2-S1=n/2[4a+4nd-2d-2a-nd+d]

=>S2-S1=n/2[2a+3nd-d]

=n/2[2a+(3n-1)d] ..(iv) by 3 we get,

=>3(S2-S1)=3n/2[2a+(3n-1)d].. (v)

From eq. (v) and (iii) we get

=>S3=3(S2-S1)

Hope it helps.

Answer 2

Answer:

Step-by-step explanation:

Let ‘a’ be the first term of the AP and ‘d’ be the common difference

S1 = (n/2)[2a + (n – 1)d] --- (1)

S2 = (2n/2)[2a + (2n – 1)d]

= n[2a + (n – 1)d] --- (2)

S3 = (3n/2)[2a + (3n – 1)d] --- (3) Consider the RHS: 3(S2 – S1) = S3

= L.H.S ∴ S3 = 3(S2 - S1) .

Thanks!