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8. Sequences and Series
hard
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
A
${S_1} + {S_3} = {S_2}$
B
${S_1} + {S_3} = 2{S_2}$
C
${S_1} + {S_2} = 2{S_3}$
D
${S_1} + {S_2} = {S_3}$
Solution
(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 ${S_1} + {S_3} = 2{S_2}$.
Standard 11
Mathematics
