Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3 (S2– S1)

Asked by Abhisek | 1 year ago |  96

##### Solution :-

Let’s take a and d to be the first term and the common difference of the A.P. respectively.

So, we have

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

Therefore,

$$s_1=\dfrac{n}{2}[2a+(n-1)d]$$

$$s_2=\dfrac{2n}{2}[2a+(n-1)d]$$

$$n[2a+(n-1)d]$$

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

Subtract S1 from S2 .

$$S_2-S_1=S_2$$

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

$$n[\dfrac{4a+4nd-2d-2a-nd+d}{2}]$$

$$n[\dfrac{2a+3nd-d}{2}]$$

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

Therefore,

S3$$\dfrac{3n}{2}[2a+(3n-1)d]$$

= 3(S2-S1)

Therefore, S3= 3(S2-S1) is proved

Answered by Pragya Singh | 1 year ago

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