Question

If S1,S2,S3 be respectively the sums of n,2n,3n terms of an AP , show that S3=3(S2-S1).

Answer 1

Answer:

It is proved below that S3=3(S2-S1).

Step-by-step explanation:

We are given that the sums of n, 2n, 3n terms of an AP is S1, S2 and S3 respectively.

We have to prove that S3 = 3(S2-S1) .

The sum of n terms of an AP formula is given by ;

        $S_n = \frac{n}{2}[2a+(n-1)d]$  ,where a = first term and d = common difference  

Now, LHS = S3 = $\frac{3n}{2}[2a + (3n-1)d]$  {as S3 represent the sum of 3n terms of AP}

RHS = 3(S2 - S1)

    S2 =  $\frac{2n}{2}[2a + (2n-1)d]$ { as S2 represent the sum of 2n terms of AP }

    S1 =   $\frac{n}{2}[2a + (n-1)d]$  { as S1 represent the sum of n terms of AP }

RHS =  3*( $\frac{2n}{2}[2a + (2n-1)d]$ - $\frac{n}{2}[2a + (n-1)d]$ )

         = $\frac{3}{2}( {2n}[2a + (2n-1)d] - n[2a + (n-1)d])$

         = $\frac{3}{2}[4na+4n^{2} d-2nd-2na-n^{2} d +nd]$

         = $\frac{3}{2}[2na+3n^{2} d-nd]$

         = $\frac{3n}{2}[2a+3nd-d]$

         =  $\frac{3n}{2}[2a + (3n-1)d]$

Hence, LHS = RHS.

So, it is proved that S3=3(S2-S1).