Question

the sum of first n terms of 3 ap are s1 , s2 and s3 respectively . the first term of each ap is 1 and common differences are 1 , 2 and 3 respectively . Prove that s1 + s3 = 2s2





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Answer 1

Given that, the sum of the first n terms of 3 APs are S1, S2, and S3.

The first term of each AP is 1.

d1 = 1
d2 = 2
d3 = 3

So,

$s_{2} = \frac{n}{2} (2a + (n - 1)d) \\ \\ = \frac{n}{2} (2 + 2n - 2) \\ \\ \frac{n}{2} (2n) \\ \\ = {n}^{2}$

$s_{1} = \frac{n}{2} (2a + (n - 1)d) \\ \\ = \frac{n}{2} (2 + n - 1) \\ \\ = \frac{n}{2} (1 + n) \\ \\ = \frac{n}{2} + \frac{ {n}^{2} }{2}$

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

Now,

$s_{1} + s_{3} = \frac{n}{2} + \frac{ {n}^{2} }{2} + \frac{3 {n}^{2} }{2} - \frac{n}{2} \\ \\ = \frac{4 {n}^{2} }{2} \\ \\ = 2 {n}^{2} = {n}^{2} \times 2 = 2 s_{2}....rhs$

Hence, proved.

Answer 2

Given that : The sum of first n terms of 3 A.P. are S1 , S2 and S3.

Also given that : The first term of each sum is 1 and common difference is 1,2 and 3 respectively.

As we know that :

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

Then,

$S1 = \frac{n}{2} [2 \times 1 + (n - 1) \times 1] \\ \\ = > S1 = \frac{n}{2} (2 + n - 1) \\ \\ = > S1 = \frac{n}{2} (1 + n) = \frac{n}{2} + \frac{ {n}^{2} }{2}$


$S2 = \frac{n}{2} [2 \times 1 + (n - 1) \times 2] \\ \\ = > S2 = \frac{n}{2} (2 + 2n - 2) = {n}^{2}$


And


$S3 = \frac{n}{2} [2 \times 1 + (n - 1) \times 3] \\ \\ = > S3 = \frac{n}{2} (2 + 3n - 3) \\ \\ = > s3 = \frac{n}{2} (3n - 1) \\ \\ = > S3 = \frac{3 {n}^{2} }{2} - \frac{n}{2}$

Now, we have to prove that : S1+S3 = 2S2

On taking LHS :

$S1 + S3 \\ \\ = > (\frac{n}{2} + \frac{ {n}^{2} }{2} ) + ( \frac{3 {n}^{2} }{2} - \frac{n}{2} ) \\ \\ = > \frac{4 {n}^{2} }{2} = 2 {n}^{2}$

On taking RHS :

$2S2 \\ \\ = > 2 \times {n}^{2} \\ \\ = > 2 {n}^{2} \\ \\ LHS = RHS \\ \\ HENCE \: PROVED$