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Let\; a_{1},a_{2},a_{3},.... be terms of an A.P. If   \frac{a_{1}+a_{2}+......+a_{p}}{a_{1}+a_{2}+......+a_{q}}=\frac{p^{2}}{q^{2}},p\neq q,then\; \frac{a_{6}}{a_{21}}\; equals

  • Option 1)

    41/11

  • Option 2)

    7/2

  • Option 3)

    2/7

  • Option 4)

    11/41

 

Answers (1)

As we learnt in 

Sum of n terms of an AP -

S_{n}= \frac{n}{2}\left [ 2a +\left ( n-1 \right )d\right ]

 

and

Sum of n terms of an AP

 

S_{n}= \frac{n}{2}\left [ a+l\right ]

- wherein

a\rightarrow first term

d\rightarrow common difference

n\rightarrow number of terms

 

Given a1 ,a2 ,a3...are in A.P and \frac{a_{1}+a_{2}+a_{3}+......+a_{p}}{a_{1}+a_{2}+a_{3}+....+a_{q}}=\frac{p^{2}}{q^{2}}

\therefore \frac{\frac{p}{2}(2a_{1}+(p-1)d)}{\frac{q}{2}(2a_{1}+(q-1)d)}=\frac{p^{2}}{q^{2}}

Put p=2\times6-1=12-1=11

      q=2\times21-1=42-1=41

\frac{2a_{1}+10d}{2a_{1}+40d}=\frac{a_{1}+5d}{a_{1}+20d}

\frac{a_{6}}{a_{21}}=\frac{11}{41}


Option 1)

41/11

this option is in-correct

Option 2)

7/2

this option is in-correct

Option 3)

2/7

this option is in-correct

Option 4)

11/41

this option is correct

Posted by

Vakul

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