Question

If S1, S2 and S4 are the sum of n, 2n and 4n terms respectively of an Ap and d is the common difference , then value of S4 - (2S1 + S2) is :
1: n^2d - nd
2: 5n^2d
3: 4n^2d
4: 6nd-2n^2d​

Answer 1

Given:

If S1, S2 and S4 are the sums of n, 2n and 4n terms respectively of an A.P. and d is the common difference

To find:

The value of S4 - (2S1 + S2)

Solution:

We have the formula for the sum of first n terms of an A.P. as:

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

So, by using the formula above we will find the value of S1, S2 & S4:

S1 = Sₙ = $\frac{n}{2}[2a + (n-1)d]$ ....... (i)

S2 = S₂ₙ = $\frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d] = 2an +2n^2d - nd$ ..... (ii)

S4 = S₄ₙ = $\frac{4n}{2}[2a + (4n-1)d] = 2n[2a + (4n-1)d] = 4an +8n^2d - 2nd$ .... (iii)

Now, we have

S4 - (2S1 + S2)

substituting the values from (i), (ii) & (iii), we get

= $4an +8n^2d - 2nd - [(2\times \frac{n}{2}[2a + (n-1)d] ) + 2an +2n^2d - nd]$

= $4an +8n^2d - 2nd - [(n}[2a + (n-1)d] ) + 2an +2n^2d - nd]$

= $4an +8n^2d - 2nd - [2an + n^2d - nd + 2an +2n^2d - nd]$

= $4an +8n^2d - 2nd - [4an + 3n^2d - 2nd]$

= $4an +8n^2d - 2nd - 4an - 3n^2d + 2nd$

= $8n^2d - 3n^2d$

= $\bold{5n^2d}$ ← option (2)

Thus, the value of S4 - (2S1 + S2) is $\underline {{5n^2d}}$.

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