Question

Answer this .. Sequenceand series​

Answer 1

Solution:-As we know that in an AP, sum of first and last term is equal to sum of second term from first and second term from last....

so,in this question

If there is an AP

a1,a2,a3,a4,.......a(2n)

then,

a1+a2n=a2+a(2n-1)=a3+a(2n-3)=......=an+a(n+1)=k(say)

so,we can see that all the numerators are equal to k,

putting this value we get

$= > \frac{k}{ \sqrt{a1} + \sqrt{a2} } + \frac{k}{ \sqrt{a2} + \sqrt{a3} } + \frac{k}{ \sqrt{a3} + \sqrt{a4} } + .... + \frac{k}{ \sqrt{an} + \sqrt{a(n + 1)} } \\ = > multiply \: and \: divide \: by \: d \: in \: all \: terms \\ = > \frac{k}{d} ( \frac{d}{ \sqrt{a1} + \sqrt{a2} } + \frac{d}{ \sqrt{a2} + \sqrt{a3} } + \frac{d}{ \sqrt{a3} + \sqrt{a4} } + .... + \frac{d}{ \sqrt{an} + \sqrt{a(n + 1)} } \\ = > \frac{k}{d} ( \frac{a2 - a1}{ \sqrt{a1} + \sqrt{a2} } + \frac{a3 - a2}{ \sqrt{a2} + \sqrt{a3} } + \frac{a4 - a3}{ \sqrt{a3 } + \sqrt{a4} } + .... + \frac{a(n + 1) - an}{ \sqrt{an} + \sqrt{a(n + 1)} } \\ using \: (a + b)(a - b) = {a}^{2} - {b}^{2} identity \: we \: get \\ = > \frac{k}{d}( \frac{( \sqrt{a2} - \sqrt{a1})( \sqrt{a2} + \sqrt{a1}) }{ \sqrt{a1} + \sqrt{a2} } + \frac{( \sqrt{a3} - \sqrt{a2})( \sqrt{a3} + \sqrt{a2} }{ \sqrt{a2} + \sqrt{3} } + \\ \frac{ (\sqrt{4} - \sqrt{a3})( \sqrt{a4} + \sqrt{a3} ) }{ \sqrt{a4} - \sqrt{a3} } + .... + \frac{ (\sqrt{an} + \sqrt{a(n + 1} )( \sqrt{a(n + 1) } - \sqrt{an}) }{ \sqrt{a(n + 1)} - \sqrt{an} } \\ cancelling \: out \: we \: get \\ = > \frac{k}{d} ( \sqrt{a2} - \sqrt{a1} + \sqrt{a3} - \sqrt{a2} + \sqrt{a4} - \sqrt{a3} + ... + \sqrt{a(n + 1)} - \sqrt{an}) \\ expanding \: the \: whole \: term \: we \: all \: terms \: will \: be \: cancelled \: out \: except \: 2 \: terms \\ = > \frac{k}{d} ( \sqrt{a(n + 1)} - \sqrt{a1}) \\ raionalizing \: num \: \: we \: get \\ = > \frac{k}{d}( \frac{a(n + 1) - a1}{ \sqrt{a(n + 1)} + \sqrt{a1} } ) \\ as \: we \: know \\ an = a1 + (n - 1)d \\ a(n + 1) = a1 + (n + 1 - 1)d = a1 + nd \\ n = \frac{a(n + 1) - a1}{d} \\ puttin \: this \: value \: of \: n \: we \: get \\ = > \frac{kn}{ \sqrt{a(n + 1)} + \sqrt{a1} } \\ also \: we \: know \\ k = a1 + a2n \\ putting \: this \: we \: get \\ = > \frac{n(a1 + a2n)}{ \sqrt{a(n + 1)} + \sqrt{a1} }$

so,

option 2) is correct one