Let A_1,A_2,A_3, ..,A_39 be 39 arithmetic means between the numbers 59 and 159.

Question

Let ${A}_{1},{A}_{2},{A}_{3},\ldots \ldots ..,{A}_{39}$ be 39 arithmetic means between the numbers 59 and 159. Then the mean of ${A}_{25},{A}_{28},{A}_{31}$ and ${A}_{36}$ is equal to :

Accepted Answer

Since there are $39$ arithmetic means between $59$ and $159$, the full arithmetic progression has

$59,{A}_{1},{A}_{2},{A}_{3},\ldots ,{A}_{39},159$

So total number of terms is

$39+2=41$

Hence, the number of intervals is

$41-1=40$

Therefore, the common difference is

$d=\frac{159-59}{40}=\frac{100}{40}=2.5$

Now the terms are:

${A}_{1}=59+2.5=61.5$

In general,

${A}_{n}=59+n(2.5)$

Now find the required terms:

${A}_{25}=59+25(2.5)=59+62.5=121.5$

${A}_{28}=59+28(2.5)=59+70=129$

${A}_{31}=59+31(2.5)=59+77.5=136.5$

${A}_{36}=59+36(2.5)=59+90=149$

Now their mean is

$\frac{{A}_{25}+{A}_{28}+{A}_{31}+{A}_{36}}{4}=\frac{121.5+129+136.5+149}{4}$

$=\frac{536}{4}=134$

So the correct answer is:

$134$

Hence, Option D is correct.

Solutions · 1