Question

Four numbers are in AP. The sum of the first three is 21 and the sum of the last three is 30. Find the numbers

Answer 1

HELLO DEAR,

Let the 4 numbers are a , a + d , a + 2d , a + 3d.

Sum of first 3 numbers = 21

a + a + d + a + 2d = 21

⇒ 3a + 3d = 21

⇒a + d = 7 ---------------( 1 )

And the sum of last three no. = 30

⇒ a + d + a + 2d + a + 3d = 30

⇒3a + 6d = 30

⇒a + 2d = 10--------------( 2 )

From----------( 1 ) & ----------( 2 )

a + d = 7

a + 2d = 10

–––––––———

-d = -3

d = 3 [ put in -------( 1 ) ]

we get,

a + 3 = 7

a = 7 - 3

a = 4 , d = 3

now, the numbers are, a = 4 , (a + d) = 7 ,(a + 2d) = 10 , (a + 3d) = 13

AP : 4 , 7 , 10 , 13

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Dear Student,

Answer:Four numbers are 4,7,10,13.

Solution:

given that AP has 4 terms

n=4

sum of first three terms = 21

sum of first three terms = 30

sum of n terms in an AP :

Sn = n/2 (2a+(n-1)d)

for first case n=3

21 = 3/2 (2a+2d)

3a+3d = 21

or a+d =7 ......eq1

for last three terms ,let we write the AP as a, a+d,a+2d ,a+3d

on addind last three term 3a+6d = 30

a+2d = 10 ....eq2

on subtracting eq 1 and 2

-d = -3

d = 3,

so a+d =7

a = 7-3 = 4

Four numbers are 4,7,10,13.

Hope it helps you.