| (i) If the ratio of sum of n terms of two A.P is \[(7n+1)\,:\,(4n+27),\] then ratio of their \[{{m}^{th}}\] terms is P. |
| (ii) Sum of n odd natural numbers is Q |
| (iii) If sum of n terms of three A.P. are |
A)
P Q R \[\frac{14m-6}{8m+23}\] \[{{n}^{2}}\] 2
B)
\[\frac{13m+6}{7m+9}\] \[{{n}^{2}}\] 5
C)
\[\frac{14m+6}{8m+23}\] \[2n+1\] 1
D)
\[\frac{7m+1}{4m+27}\] \[2n-1\] 3
Correct Answer: A
Solution :
(i) Let \[{{S}_{n}}\] be the sum of n terms of 1st A.P. and \[S{{'}_{n}}\] be sum of n terms of 2nd A.P. According to question, \[\frac{{{S}_{n}}}{S{{'}_{n}}}=\frac{7n+1}{4n+27}\] \[\frac{\frac{n}{2}\left[ 2{{a}_{1}}+\left( n-1 \right){{d}_{1}} \right]}{\frac{n}{2}\left[ 2{{a}_{2}}+\left( n-1 \right){{d}_{2}} \right]}=\frac{7n+1}{4n+27}\] \[\Rightarrow \] \[\frac{2{{a}_{1}}+\left( n-1 \right){{d}_{1}}}{2{{a}_{2}}\left( n-1 \right){{d}_{2}}}=\frac{7n+1}{4n+27}\] Put \[n=(2m-1)\] in above equation, we get \[\frac{2{{a}_{1}}+\left( 2m-1-1 \right){{d}_{1}}}{2{{a}_{2}}+\left( 2m-1-1 \right){{d}_{2}}}=\frac{7\left( 2m-1 \right)+1}{4\left( 2m-1 \right)+27}\] \[\Rightarrow \] \[\frac{\left[ {{a}_{1}}+\left( m-1 \right){{d}_{1}} \right]}{\left[ {{a}_{2}}+\left( m-1 \right){{d}_{2}} \right]}=\frac{14m-6}{8m+23}\] \[\Rightarrow \] \[\frac{{{a}_{m}}}{{{a}_{m}}}=\frac{14m-6}{8m+23}\] (ii) A.P. of odd n natural numbers is 1, 3, 5, ............ \[{{S}_{n}}=\frac{n}{2}\left[ 2(1)+\left( n-1 \right)\left( 2 \right) \right]=\frac{n}{2}\times 2n={{n}^{2}}\] (iii) We have, \[{{S}_{1}}=\frac{n}{2}\left[ 2(1)+\left( n-1 \right)\left( 1 \right) \right]=\frac{n}{2}\left[ n+1 \right]\] \[{{S}_{2}}=\frac{n}{2}\left[ 2(1)+\left( n-1 \right)\left( 2 \right) \right]=\frac{n}{2}\times 2n={{n}^{2}}\] \[{{S}_{3}}=\frac{n}{2}\left[ 2(1)+\left( n-1 \right)\left( 3 \right) \right]=\frac{n}{2}\left[ 3n-1 \right]\] Now, \[\frac{{{S}_{1}}+{{S}_{3}}}{{{S}_{2}}}=2\]
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