Question

4. The sum of first n terms of three A.P are S1 ,S2 and S3 respectively. The first term of each A.P is 1 and their common difference are 1, 2 and 3 respectively .Prove that S1 +S3= S2​

Answer 1

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The sum of first n terms of three A.P are S1 ,S2 and S3 respectively. The first term of each A.P is 1 and their common difference are 1, 2 and 3 respectively .Prove that S1 +S3= S2

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$<font color=red><marquee behavior=alternate>Solution</marquee></font>$

For the first AP series....

First term(a)=1

difference (d1)=1

now...

$S1(n)=\frac{n(2a + (n - 1)d1)}{2} \\ \: \: \: \: \: = \frac{n(2 \times 1 + (n - 1)1)}{2} \\\: \: \: \: \: = \frac{n(n + 1)}{2} \\ \: \: \: \: \: = \frac{n {}^{2} + n}{2}$

For the second AP series....

First term(a)=1

difference (d2)=2

now...

$S2(n)=\frac{n(2a + (n - 1)d2)}{2} \\\: \: \: \: \: = \frac{n(2 \times 1+ (n - 1)2)}{2} \\\: \: \: \: \: = \frac{n(2n )}{2} \\\: \: \: \: \: = n {}^{2}$

For the third AP series....

First term(a)=1

difference (d3)=3

now...

$S3(n)=\frac{n(2a + (n - 1)d3)}{2} \\\: \: \: \: \: = \frac{n(2 \times 1 + (n - 1)3)}{2} \\\: \: \: \: \: = \frac{n(n - 1)}{2} \\ \: \: \: \: \: = \frac{n {}^{2} - n}{2}$

Now.....

$S1(n)+S3(n)=\frac{n {}^{2} + n}{2}+ \frac{n {}^{2} - n}{2}\\\: \: \: \: \: = \frac{n {}^{2} - n+n {}^{2} +n}{2} \\\: \: \: \: \: =n{}^{2}\\\: \: \: \: \: =S2(n)$

therefore.....

$\:\:\:\: \: \: \: \: \: \large\mathfrak{\underline{\huge\mathcal{\bf{\boxed{\boxed{\huge\mathcal{S1+S3=S2}}}}}}}$

$<font color=green><marquee direction="Right">hope this help you$

Answer 2

Step-by-step explanation:

Sum of n terms of first progression :

$</p><p>S_{1} = \frac{n}{2} </p><p>[2.1 + (n - 1).1]$

$⟹ \frac{n}{2} (2 + n - 1)$

$⟹ \frac{n(n + 1)}{2}$

Sum of n terms of second progression :

$S _{2} = \frac{n}{2} [2.1 + (n - 1).2]$

$⟹ \frac{n}{2} \: [2 + (n - 1).2]$

$⟹n(1 + n - 1)$

$⟹ {n}^{2}$

Sum of n terms of third progression :

$S _{3} = \frac{n}{2} [2.1 + (n - 1).3]$

$⟹ \frac{n}{2} (2 + 3n - 3)$

$⟹ \frac{n}{2} (3n - 1)$

$∴ \: \: \: \: S_{1} + S_{3} = \frac{n(n + 1)}{2} + \frac{n(3n - 1)}{2}$

$⟹ \frac{n}{2} \times (n + 1 + 3n - 1)$

$⟹ \frac{n}{2} \times 4n = 2 {n}^{2}$

$⟹2S _{2}, \: \: \: \: [∴ S _{2} = n²]$

Proved.