Let l, m and n are three distinct numbers in arithmetic progression. Also l^2, m^2 and n^2 are in geometric progression and l+m+n=3. If l

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Let l, m and n are three distinct numbers in arithmetic progression. Also l^2, m^2 and n^2 are in geometric progression and l+m+n=3. If l<m<n, then n is equal to

MARKS Content Team · Accepted Answer

Correct answer: (C) 1+√2. l,m,n are in A.P. ⇒2m=l+n Given, l+m+n=3⇒m=1 Let, l=1-d, n=1+d, d>0 Now, m^4=l^2n^2⇒m^2=-ln or ln So 1+d1-d=±1 ⇒d=0 or √2 (d=0 is rejected) Hence, n=1+√2

Let $l, m$ and $n$ are three distinct numbers in arithmetic progression. Also $l^{2}, m^{2}$ and $n^{2}$ are in geometric progression and $l + m + n = 3$ . If $l < m < n,$ then $n$ is equal to

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More Sequences and Series Questions

Let l, m and n are three distinct numbers in arithmetic progression. Also l2, m2 and n2 are in geometric progression and l+m+n=3. If l<m<n, then n is equal to

Solution

More Sequences and Series Questions

Let $l, m$ and $n$ are three distinct numbers in arithmetic progression. Also $l^{2}, m^{2}$ and $n^{2}$ are in geometric progression and $l + m + n = 3$ . If $l < m < n,$ then $n$ is equal to