Question

If the sum ofn, 2n, 3n terms of an arithmetic progression are s,, S, S, respectively
then prove that s3 = 3 (s,-s,).​

Answer 1

Correct Question :-

The sum of n , 2n , 3n terms of an AP are

S₁ , S₂ S₃ respectively , then prove that

S₃= 3(S₂ - S₁).

Answer:-

Given:

Sum of first n terms of an AP = S₁

Sum of first 2n terms = S₂

Sum of first 3n terms = S₃

We know that,

Sum of first n terms (Sₙ) = n/2 [ 2a + (n - 1)d ]

So,

★ S₁ = n/2 [ 2a + (n - 1)d ] -- (1)

★ S₂ = 2n/2 [ 2a + (2n - 1)d ] -- (2)

★ S₃ = 3n/2 [ 2a + (3n - 1)d ] -- (3)

Now,

We have to prove that:

S₃= 3 (S₂ - S₁)

Putting the values from equations (1) , (2) , (3) we get,

$\implies \sf \: \frac{3n}{2} [2a + (3n - 1)d] = 3 \bigg( \frac{2n}{2} [2a + (2n - 1)d] - \bigg\{ \frac{n}{2} [2a + (n - 1)d ] \bigg\} \bigg) \\ \ \\ \ \implies \sf \: \frac{3n}{2}(2a + 3nd - d) = 3 \bigg( n(2a + 2nd - d) - \bigg \{ \frac{n}{2} (2a + nd - d) \bigg \} \bigg) \\ \\ \\ \implies \sf \: \frac{6an + 9 {n}^{2}d - 3nd }{2} = 3 \bigg( \frac{2 \times n(2a + 2nd - d) - n(2a + nd - d)}{2} \bigg) \\ \\ \\ \implies \sf \:\: \frac{6an + 9 {n}^{2} d - 3nd}{2} = 3 \bigg( \frac{4an +4{n}^{2}d -2 nd - 2an - {n}^{2} d + nd }{2} \bigg) \\ \\ \\ \implies \sf \:\: \frac{6an + 9 {n}^{2} d - 3nd}{2} = 3 \bigg( \frac{2an +3 {n}^{2} d - nd }{2} \bigg) \\ \\ \\ \implies \sf \:\: \frac{6an + 9 {n}^{2} d - 3nd}{2} = \frac{6an + 9 {n}^{2} d - 3nd}{2}$

Hence Proved.

Answer 2

$\large{\bf{\underline{Corrrect \: Question:-}}}$

If the sum of n, 2n and 3n terms of an A.P. are $\sf S_1 , S_2 , S_3$ respectively. Then, prove that $\sf S_3 = 3(S_2 - S_1)$

$\large{\bf{\underline{AnSwer:-}}}$

$\rule{278}{2}$

$\large{\bf{\underline{ \green{Given:-}}}}$

❐ Sum of n, 2n and 3n terms are $\sf S_1 , S_2 , S_3$ respectively.

$\large{\bf{\underline{ \green{To \: Prove:-}}}}$

❐ $\sf S_3 = 3(S_2 - S_1)$

$\large{\bf{\underline{ \green{Solution:-}}}}$

Here, Let (a) be the first term and (d) be the common difference of the given A.P.

Now, using

$\large{ \red{\underline{ \pink{\boxed{\sf S_n = \dfrac{n}{2} \bigg\lgroup {2a + (n-1)d} \bigg\rgroup}}}}}$

Now,

Sum of n terms = $\sf S_1$

⇰$\: \: \: \sf S_1 = \dfrac{n}{2} \bigg\lgroup {2a + (n-1)d} \bigg\rgroup \: \: ........ \mathcal{(i)}$

Sum of 2n terms = $\sf S_2$

⇰$\: \: \: \sf S_2 = \dfrac{2n}{2} \bigg\lgroup {2a + (2n-1)d} \bigg\rgroup \: \: ........ \mathcal{(ii)}$

Sum of 3n terms = $\sf S_3$

⇰$\: \: \: \sf S_3 = \dfrac{3n}{2} \bigg\lgroup {2a + (3n-1)d} \bigg\rgroup \: \: ........ \mathcal{(iii)}$

$\sf \red{\bigstar} \underline{Subtract \: eq.(i) \: from \: eq.(ii)} \blue{\bigstar}$

$\sf\implies S_2 - S_1 = \dfrac{2n}{2} \bigg\lgroup {2a + (2n-1)d} \bigg\rgroup - \dfrac{n}{2} \bigg\lgroup {2a + (n-1)d} \bigg\rgroup$

$\sf\implies S_2 - S_1 = \dfrac{n}{2} \bigg\lgroup {2 \bigg(2a + (2n-1)d} \bigg) - \bigg(2a + (n - 1)d \bigg) \bigg\rgroup$

$\sf\implies S_2 - S_1 = \dfrac{n}{2} \bigg\lgroup {2a + (3n - 1)d } \bigg\rgroup$

$\sf\implies 3(S_2 - S_1)= \dfrac{3n}{2} \bigg\lgroup {2a + (3n - 1)d } \bigg\rgroup = S_3$

$\sf\therefore S_3 = 3(S_2 - S_1)$

Hence, $\sf S_3 = 3(S_2 - S_1)$

$\bold{ \pink{Hence \: Proved}}$