If Sn denotes the sum of first n terms of an A-P- such that S mSn - m 2 n 2- thenaman -a 2 m -12 n -1b 2 m -12 n -1c m -1 n -1d m -1 n -1

(b) 2 #160;m 12 #160;n 1 SmSn=m2n2 #8658;m22a+m 1dn22a+n 1d=m2n2 #8658;2a+m 1d2a+n 1d=mn #160; #8658;2an+ndm nd=2am+nmd md #8658;2an 2am nd+md=0 ...

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Standard XII

Mathematics

Sum of n Terms

If Sn denotes...

Question

If S_n denotes the sum of first n terms of an A.P. < a_n > such that $\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}, then \frac{a_{m}}{a_{n}} =$
(a) $\frac{2 m + 1}{2 n + 1}$

(b) $\frac{2 m - 1}{2 n - 1}$

(c) $\frac{m - 1}{n - 1}$

(d) $\frac{m + 1}{n + 1}$

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Solution

(b) $\frac{2 m - 1}{2 n - 1}$

$\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}} \Rightarrow \frac{\frac{m}{2} \{2 a + (m - 1) d\}}{\frac{n}{2} \{2 a + (n - 1) d\}} = \frac{m^{2}}{n^{2}} \Rightarrow \frac{\{2 a + (m - 1) d\}}{\{2 a + (n - 1) d\}} = \frac{m}{n} \Rightarrow 2 a n + n d m - n d = 2 a m + n m d - m d \Rightarrow 2 a n - 2 a m - n d + m d = 0 \Rightarrow 2 a (n - m) - d (n - m) = 0 \Rightarrow 2 a (n - m) = d (n - m) \Rightarrow d = 2 a . . . . . (1)$

$Ratio of \frac{a_{m}}{a_{n}} = \frac{a + (m - 1) d}{a + (n - 1) d} \Rightarrow \frac{a_{m}}{a_{n}} = \frac{a + (m -) 2 a}{a + (n -) 2 a} From (1) = \frac{a + 2 a m - 2 a}{a + 2 a n - 2 a} = \frac{2 a m - a}{2 a n - a} = \frac{2 m - 1}{2 n - 1}$

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If Sn denotes the sum of first n terms of an A.P. < an > such that SmSn=m2n2, thenaman= (a) 2 m+12 n+1 (b) 2 m-12 n-1 (c) m-1n-1 (d) m+1n+1

(b) 2 m-12 n-1 SmSn=m2n2⇒m22a+m-1dn22a+n-1d=m2n2⇒2a+m-1d2a+n-1d=mn ⇒2an+ndm-nd=2am+nmd-md⇒2an-2am-nd+md=0⇒2an-m-dn-m=0⇒2an-m=dn-m⇒d=2a .....1 Ratio of aman=a+m-1da+n-1d⇒ aman=a+m-2aa+n-2a From (1) = a+2am-2aa+2an-2a =2am-a2an-a = 2m-12n-1

If S_n denotes the sum of first n terms of an A.P. < a_n > such that $\frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}}, then \frac{a_{m}}{a_{n}} =$
(a) $\frac{2 m + 1}{2 n + 1}$

(b) $\frac{2 m - 1}{2 n - 1}$

(c) $\frac{m - 1}{n - 1}$

(d) $\frac{m + 1}{n + 1}$