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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

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Aadil

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