Question

An A.P consists of three terms whole sum is 15 and sum of their squares of extremes is 58 find the first three terms of AP and also find the sum of first 50 teams of AP

Answer 1

let first term be a

& common difference be d

three terms will be a-d,a,a+d

sum of these terms would be = 15

3a=15

a=5

sum of their extremes would be 58

(a-d)²+(a+d)²=58

2(a²+d²)=58

(a²+d²)=29

(5²+d²)=29

d²=29-25=4

d=+-2

three terms would be if d=+2

3,5,7

three terms of ap, if d = -2

7,5,3

sum of 50 terms would be n/2[2a+(n-1)d]

n = 50 , a = 3 , d=+2

sum of 50  terms = 25[6+49(2)] = 2600

n= 50 , a=7 , d= -2

sum of 50 terms = 25[14+49(-2)]=-2100