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**If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are**

A. 5, 10, 15, 20
B. 4, 10, 16, 22
C. 3, 7, 11, 15
D. None of these
**Answer: Option A**

## Show Answer

Solution(By Apex Team)

4 numbers are in A.P.
Let the numbers be
a – 3d, a – d, a + d, a + 3d
Where a is the first term and 2d is the common difference
Now their sum = 50
a – 3d + a – d + a + d + a + 3d = 50
and greatest number is 4 times the least number
a + 3d = 4 (a – 3d)
a + 3d = 4a – 12d
4a – a = 3d + 12d
⇒ 3a = 15d
$\begin{aligned}&\Rightarrow a=\frac{15d}{3}=5d\\
&\Rightarrow\frac{25}{2}=5d\\
&\Rightarrow d=\frac{25}{2\times5}\\
&\Rightarrow d=\frac{5}{2}\end{aligned}$
$\begin{aligned}&\therefore\text{ Numbers are }\\
&\frac{25}{2}-3\times\frac{5}{2},\ \frac{25}{2}-\frac{5}{2},\ \frac{25}{2}+\frac{5}{2},\ \frac{25}{2}+3\times\frac{5}{2}\\
&\Rightarrow\frac{10}{2},\ \frac{20}{2},\ \frac{30}{2},\ \frac{40}{2}\\
&\Rightarrow5,\ 10,\ 15,\ 20\end{aligned}$

## Related Questions On Progressions

### How many terms are there in 20, 25, 30 . . . . . . 140?

A. 22B. 25

C. 23

D. 24

### Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.

A. 5B. 6

C. 4

D. 3

### Find the 15th term of the sequence 20, 15, 10 . . .

A. -45B. -55

C. -50

D. 0

### The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is

A. 600B. 765

C. 640

D. 680