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B
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Difficulty:
95%
(hard)
Question Stats:
49% (03:36) correct
51%
(03:23)
wrong
based on 43
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In a particular sequence, the general term \(a_n\) is defined as \(a_n\) = 2*\(a_i \)+ 3*\(a_j\) for n > 2, where i = n-1 and j=n-2.
If \(a_5\) is 17 more than twice the first term and \(a_3\) is two-fifth of the fourth term, what is the value of \(a_2\) ?
(A) 1/2
(B) 3/8
(C) 2
(D) 8/3
(E) 17
If \(a_5\) is 17 more than twice the first term and \(a_3\) is two-fifth of the fourth term, what is the value of \(a_2\) ?
(A) 1/2
(B) 3/8
(C) 2
(D) 8/3
(E) 17
13 Jun 2025, 15:09
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siddhantvarma
Note :
\(a_{5} = 2a_{4}+3a_{3}\)...(i)
\(a_{4} = 2a_{3}+3a_{2}\)...(ii)
\(a_{3} = 2a_{2}+3a_{1} \)...(iii)
\(a_{5}-2a_{1} =17\)...(iv)
Reducing (iv) till we get it in terms of \(a_{2}\) and \(a_{1}\)
\(2a{_4}+3a_{3}-2a_{1}=17\)...Substituting the value of \(a_{5}\) in (iv)
\(2(2a_{3}+3a_{2})+3(2a_2+3a_{1})-2a_{1}=17\)...Substituting the value of \(a_{4}\) and \(a_{3}\)
Similarly further substituting the value of \(a_{3}\) above and simplifying
\(19a_{1}+20a_{2}=17\)...(v)
Also given
\( a_{3} = \frac{2}{5}a_{4}\) ...(vi)
Like before again reducing (vi) till we get it in terms of \(a_{1}\) and \(a_{2}\)
\( 2a_{2}+3a_{1} = \frac{2}{5} ( 2a_{3}+3a_{2}) \)... Substituting the value of \(a_{3}\) and \(a_{4}\)
...again further substituting the value of \(a_{3}\) above and simplifying.
\(3a_{1}= 4a_{2}\)
\(a_{1}= \frac{4}{3}a_{2}\)
\(19* \frac{4}{3}a_{2} + 20a_{2}=17\)... substituting the value of \(a_{1}\) in (v)
\( a_{2}= \frac{51}{136} = \frac{3}{8} \)
Ans B
Hope it helped.
