Question

The sum of first n terms of three AP's are S1, S2 and S3. The first term of each is 5 and their common differences are 2,4 and 6 respectively. Prove that S1+S3=2S2.

Answer 1

To prove:

$\mathrm{S}_{1}+\mathrm{S}_{3}=2 \mathrm{S}_{2}$

Solution:

Given: The sum of first ‘n’ terms of three AP's are $\mathrm{S}_{1}, \mathrm{S}_{2} \text { and } \mathrm{S}_{3}$. The ‘first term’ of each is 5 and their common differences are 2, 4 and 6 respectively.  

Formula of summation of an A.P

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

whereas  = Summation till n terms.

a = First term of sequence

d = Common Difference

Now, using this formula, we get  

$S_{1}=\frac{\left(n+n^{2}\right)}{2}$

$S_{2}=n^{2}$

$S_{3}=\frac{\left(3 n^{2}-n\right)}{2}$

Thus, we need to add $S_{1}\ and\ S_{3}$ to prove that $\mathrm{S}_{1}+\mathrm{S}_{3}=2 \mathrm{S}_{2}$

$\therefore S_{1}+S_{3}=\frac{\left(n+n^{2}\right)}{2}+\frac{\left(3 n^{2}-n\right)}{2}$

$=\frac{n^{2}+3 n^{2}}{2}$

$=\frac{4 n^{2}}{2}$

$=2 n^{2}$

$=2\left(S_{2}\right).$                              

Thus, $\mathrm{S}_{1}+\mathrm{S}_{3}=2 \mathrm{S}_{2}$.

Hence proved.

Answer 2

Answer:

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