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

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(b) We have ${a_1} = {a_2} = {a_3} = 1$ and ${d_1} = 1,\;{d_2} = 2,\;{d_3} = 3$.

Therefore, ${S_1} = \frac{n}{2}(n + 1)$ ......(i)

${S_2} = \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) We have ${a_1} = {a_2} = {a_3} = 1$ and ${d_1} = 1,\;{d_2} = 2,\;{d_3} = 3$.

Therefore, ${S_1} = \frac{n}{2}(n + 1)$ ......(i)

${S_2} = \frac{n}{2}[2n]$ ......(ii)

${S_3} = \frac{n}{2}[3n - 1]$ ......(iii)

Adding (i) and (iii),

${S_1} + {S_3} = \frac{n}{2}[(n + 1) + (3n - 1)] $

$= 2\left[ {\frac{n}{2}(2n)} \right] = 2{S_2}$

Hence correct relation ${S_1} + {S_3} = 2{S_2}$.

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

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