Question

If the sum of 1st n, 2n and 3n terms of an AP are S1,S2,S3 respectively, then prove that S3 = 3(S2-S1)​

Answer 1

☸ Sᴏʟᴜᴛɪᴏɴ ☸

Sn = n/2 [2a + (n - 1)d]

⟼ S3 = S3n = 3n/2 [2a + (3n - 1)d]

⟼ S2 = S2n = 2n/2 [2a + (2n - 1)d]

⟼ S1 = Sn = n/2 [2a + (n - 1)d]

⟼ S3 = 3n/3 [2a +n,(3n - 1)d] .…...……..❶

⛯ Now, Find the value of S2 - S1

→ ❴ 2n/2 [2a + (2n - 1) d] ❵ - ❴ n/2 [2a + (n - 1) d ❵

→ ❴ 2n[2a + (2n - 1) d] / 2 ❵ - ❴ n[2a + (n - 1) d] /2 ❵

→ ❴ 4na + 4n²d - 2nd - 2na - n²d + nd / 2 ❵

→ ❴ 2na + 3n²d - nd / 2 ❵

→ ❴ n(2a + 3d - d) / 2 ❵

→ n[2a + (3n - 1) d] / 2

→ n/2 [ 2a + (3n - 1) d ]

✍ Here, we have

↪ S2 - S1 = n/2 [2a + (3n - 1) d]

✒ Now, Multiplying both sides by 3

↠ 3(S2 - S1) = 3[n/2{2a + (3n - 1)d}]

↠ 3(S2 - S1) = 3n/2 [2a + (3n - 1)d].…...……..❷

❥ From equation ❶ & ❷, we can write as :

S3 = 3(S2 - S1)

★ Hence Proved ★

Answer 2

Step-by-step explanation:

Given :-

• The sum of n,2n and 3n terms of an AP are $\sf{S_1,S_2,S_3}$ respectively.

To prove :-

• $\sf{S_3=3(S_2-S_1)}$

Solution :-

Consider,

• 1st term = a
• Common difference = d

The sum of n,2n and 3n terms of an AP are $\sf{S_1,S_2,S_3}$ respectively.

Then,

★$\sf{S_1=S_n=\dfrac{n}{2}[2a+(n-1)d]}$

★$\sf{S_2=S_{2n}=\dfrac{2n}{2}[2a+(2n-1)d]}$

★$\sf{S_3=S_{3n}=\dfrac{3n}{2}[2a+(3n-1)d]}$

Taking R.H.S,

$\sf{3(S_2-S_1)}$

• Put values.

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

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

$\to\sf{\dfrac{3n}{2}\bigg(2[2a+2nd-d]-[2a+nd-d]\bigg)}$

$\to\sf{\dfrac{3n}{2}\bigg(4a+4nd-2d-2a-nd+d\bigg)}$

$\to\sf{\dfrac{3n}{2}\bigg(2a+3nd-d\bigg)}$

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

$\to\sf{S_{3n}}$

$\to\sf{S_3}$

L.H.S = R.H.S [ Proved ]