Question

Find four consecutive terms in A.P. whose sum is 20 and sum of 3rd and 4th term is 1.

Answer 1

Answer:

Step-by-step explanation:

a-3d+a-d+a+d+a+3d=20

4a=20

a=5

a+2d+a+3d=1

5+5+5d=1

5d=-9

d=-9/5

So the terms after substitution of a & b are:

-2/5,16/5,34/5,52/5

Answer 2

Answer: 47/4,29/4,11/4,-7/4

Step-by-step explanation:

Let a be the first ter and d be the common difference

So four consecutive terms be a,a+d,a+2d,a+3d.

Given sum of four terms is 20

So...4a+6d=20

=>2a+3d=10 - equ1

Also given sum of 3rd and 4rth term is 1

So...2a+5d=1 - equ3

By solving equ 1 and 2..we will get

a=47/4 and d= -9/2

So the terms should be 47/4,29/4,11/4, -7/4

Hope this answer is correct and hope it might help you...☺