The sums of n terms of three A.P.'s whose fir | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.Given $a = 1$ for all three $A.P.'s$ and common differences $d_1

Vedclass · Accepted Answer

${S_1} + {S_3} = 2{S_2}$

Vedclass · Answer

(B) The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.Given $a = 1$ for all three $A.P.'s$ and common differences $d_1 = 1, d_2 = 2, d_3 = 3$.For $S_1$: $S_1 = \frac{n}{2}[2(1) + (n - 1)1] = \frac{n}{2}[n + 1]$.For $S_2$: $S_2 = \frac{n}{2}[2(1) + (n - 1)2] = \frac{n}{2}[2n] = n^2$.For $S_3$: $S_3 = \frac{n}{2}[2(1) + (n - 1)3] = \frac{n}{2}[3n - 1]$.Now,calculating $S_1 + S_3$:$S_1 + S_3 = \frac{n}{2}[n + 1 + 3n - 1] = \frac{n}{2}[4n] = 2n^2$.Since $S_2 = n^2$,we have $2S_2 = 2n^2$.Therefore,$S_1 + S_3 = 2S_2$.

The sums of $n$ terms of three $A.P.'s$ whose first term is $1$ and common differences are $1, 2, 3$ are ${S_1}, {S_2}, {S_3}$ respectively. The true relation is

Similar Questions

The sum of the series $3 + 33 + 333 + \dots + n$ terms is

If the sum of first $n$ terms of an $A.P.$ is equal to the sum of its first $m$ terms,$(m \ne n)$,then the sum of its first $(m + n)$ terms will be

$(1^{2}+2^{2}+3^{2}+\cdots+10^{2})$ is equal to:

If the $m^{th}$ terms of the series $63 + 65 + 67 + 69 + \dots$ and $3 + 10 + 17 + 24 + \dots$ are equal,then $m = $

If the $p^{th}$ term of an $A.P.$ is $q$ and the $q^{th}$ term is $p,$ then its $r^{th}$ term is

Vedclass Test Series

Exam Paper Generator

Online Exam Module