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 Let the sum of the first three terms of an A.P. be 39 and the sum of its last four terms be 178. 

If the first term of this A.P. is 10, then the median of the A.P. is :

  • Option 1)

     26.5

  • Option 2)

     28

  • Option 3)

    29.5

  • Option 4)

    31

 

Answers (1)

best_answer

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

 

Arithmetic mean of n numbers -

A= \frac{1}{n}\left ( a_{1}+a_{2}+a_{3}- - - - - a_{n} \right )

 

- wherein

a_{1}, a_{2}, a_{3}- - - - - a_{n} are the n numbers

 

 Let the AP is a, a+d, a+2d ...a+\left ( n-4 \right )d, a+\left ( n-3 \right )d, a+\left ( n-2 \right )d, a+\left ( n-1 \right )d

given a+a+d+a+2d=39

3a+3d=39

a+d=13

d=13-10

 d=3\left [ \because a=10 given \right ]

Now a+\left ( n-4 \right )d+a+\left ( n-3 \right )d+a+\left ( n-2 \right )d+a\left ( n-1 \right )d

\Rightarrow 4a+d\left ( 4n-10 \right )=178

\therefore n=14 \left [ put\ a=10 \right ]

                   \left [ d=3 \right ]

\therefore Series is 10, 13, 16 - - - - -  upto 14 terms

S_{14}=\frac{14}{2}\left [ 2\times 10\dotplus 13\times 3 \right ]=7\left [ 20+39 \right ]

\therefore mean= \frac{s_{14}}{14}

=\frac{7}{14}\left ( 59 \right )

=\frac{59}{2}

=29.5


Option 1)

 26.5

Incorrect option

Option 2)

 28

Incorrect option

Option 3)

29.5

Correct option

Option 4)

31

Incorrect option

Posted by

Aadil

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