Let S1, S2, S3 be the respective sums of first n, 2 n and 3 n terms of the same arithmetic progression with a as the first term and d as the common difference. If R=S3-S2-S1, then R depends on

Question

Let S1, S2, S3 be the respective sums of first n, 2 n and 3 n terms of the same arithmetic progression with a as the first term and d as the common difference. If R=S3-S2-S1, then R depends on (A) a and d (B) d and n (C) a and n (D) a, d and n

Tardigrade Admin · Accepted Answer

R =S3-S2-S1 =(3 n/2)(2 a+(3 n-1) d)-(2 n/2)(2 a+(2 n-1) d) -(n/2)(2 a+(n-1) d) =2 a((3 n/2)-(2 n/2)-(n/2))+(n d/2)(3(3 n-1) -2(2 n-1)-1(n-1)) =(n d/2)(4 n)=2 n2 d

Q. Let $S_{1}, S_{2}, S_{3}$ be the respective sums of first $n, 2 n$ and $3 n$ terms of the same arithmetic progression with $a$ as the first term and $d$ as the common difference. If $R = S_{3} - S_{2} - S_{1}$ , then $R$ depends on

$a$ and $d$

$d$ and $n$

$a$ and $n$

$a, d$ and $n$

$R = S_{3} - S_{2} - S_{1}$
$= \frac{3 n}{2} (2 a + (3 n - 1) d) - \frac{2 n}{2} (2 a + (2 n - 1) d)$
$- \frac{n}{2} (2 a + (n - 1) d)$
$= 2 a (\frac{3 n}{2} - \frac{2 n}{2} - \frac{n}{2}) + \frac{n d}{2} (3 (3 n - 1)$
$- 2 (2 n - 1) - 1 (n - 1))$
$= \frac{n d}{2} (4 n) = 2 n^{2} d$

Q. Let S1​,S2​,S3​ be the respective sums of first n,2n and 3n terms of the same arithmetic progression with a as the first term and d as the common difference. If R=S3​−S2​−S1​, then R depends on

a and d

d and n

a and n

a,d and n