The sum of first n, 2nn, 2nn, 2n, and 3n3n3n terms of an AP is S1, S2S1, S2S1, S2, and S3S3S3 respectively. Prove that S3=3(S2−S1).S3=3(S2−S1).S3=3(S2−S1).
Answer:
-
We are told that
S1S1S1 = Sum of first nnn terms
S2S2S2 = Sum of first 2n2n2n terms
S3S3S3 = Sum of first 3n3n3n terms
- We know that the sum of first nnn terms of an AP is given by
Sn=n2(2a+(n−1)d),Sn=n2(2a+(n−1)d),Sn=n2(2a+(n−1)d), where aaa is the first term and nnn is the number of terms in the AP.
Therefore, we have [Math Processing Error] - Now, [Math Processing Error]
- Thus, S3=3(S2−S1).S3=3(S2−S1).