struggling wid basic maths_arthematic progression...

willbeatthegmat wrote:The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP?

WIKI has a perfect explanation, which you fill find very useful:

https://en.wikipedia.org/wiki/Arithmetic_progression

4th term: a1+3d
8th term: a1+7d

2a1+10d = 24 I

6th term: a1+5d
10th term: a1+9d

2a1+14d=44 II

II - I :

4d=20, d=5 and solve it for a1, you will get a1= -13

a1=-13
a2=-8
a3=-3

In an AP

an = a1+ (n-1) d

where a - first term n- number of terms and d-> common difference

Given a4+a8 = 24
a6+a10 = 44

a4 = a1+3d
a8 = a1+7d
a6 = a1+5d
a10 = a1+9d

a1+3d+a1+7d = 24 -> 2a1+10d = 24->a1+5d=12 (1)

a1+5d+a1+9d = 44-> 2a1+14d = 44->a1+7d=22 (2)

Solve for a1 and d from 1 and 2

d=5
a1=-13
a2=a1+d = -13+5 = -8
a3= a1+2d = -13+2(5) = -3


Hope this helps.

Whats the answer given?

AP = arithmetic progression

formula for AP is the nth term = a + (n-1)d, where a=first term



4th term: a + (4-1)d = a+3d

8th term: a + (8-1)d = a+7d

a+3d+a+7d=24
2a + 10d = 24
a + 5d = 12 (i)

6th term: a + (6-1)d = a+5d

10th term: a + (10-1)d = a+9d

a+5d+a+9d=44
2a + 14d = 44
a + 7d = 22 (ii)

subtract the first from the second to get:

2d = 10
d=5

substitute d back into either i or ii

a + 7(5) = 22
a = -13


nth term = a + (n-1)d

a2 = -13 + (2-1)5
a2 = -8

a3 = -13 + (3-1)5
a3 = -3

First three terms are -13, -8, -3

cramya wrote:In an AP

an = a1+ (n-1) d

where a - first term n- number of terms and d-> common difference

Given a4+a8 = 24
a6+a10 = 44

a4 = a1+3d
a8 = a1+7d
a6 = a1+5d
a10 = a1+9d

a1+3d+a1+7d = 24 -> 2a1+10d = 24->a1+5d=12 (1)

a1+5d+a1+9d = 44-> 2a1+14d = 44->a1+7d=22 (2)

Solve for a1 and d from 1 and 2

d=5
a1=-13
a2=a1+d = -13+5 = -8
a3= a1+2d = -13+2(5) = -3


Hope this helps.

Whats the answer given?

Beat me too it! Nice work cramya and logitech!

ans is -13,-8,-3.....too gud guys..u made it look preety simple