Answer:
Rs 159
Explanation:
1000+140 = 1140
let first installment be a
second will be a-10, third will be a-20
hence we have
a+(a-10)+(a-20)+...........10 TERMS
CLEARLY THIS IS AN ARITHMETIC PROGRESSION
Sn = (n/2){2a+(n-1)d}
here Sn = 1140 n = 10 , d = a-10-a = -10
1140 = (10/2){2a +(10-1)(-10)}
1140= 5{2a+(9)(-10)}
1140= 5{2a-90}
5{2a-90} = 1140
2a-90 = 1140/5
2a-90 = 228
2a = 228+90
2a = 318
a = 318/2
a= 159