Please solve this problem

Let the X is a amount saved in year 1, so in year 2 it will be x 100,in year 3 it will be x 200,in year 4 it will be x 300 and so on. .In year 10 it will be x 900. Sum of all there is. . 10x 4500 which is equal to 16500 so X is equal to 1200. . .

Let the X is a amount saved in year 1, so in year 2 it will be x 100,in year 3 it will be x 200,in year 4 it will be x 300 and so on. .In year 10 it will be x 900. Sum of all there is. . 10x 4500 which is equal to 16500 so X is equal to 1200. . .

i don't understand.

1. Ans.

suppose amount saved in first year is Rs a.

hence amt saved in next years will be a+100, a+200,a+300 and so on.............

and also we have Sum of nth term = n/2{2a+(n-1)d}

Sn= 10/2{2a+(10-1)100}

or, 16500= 5(2a+900)

or,2a+900= 16500/5

or, 2a+900= 3300

or, 2a= 3300-900

or, a= 2400/2 = 1200

hence saving in first year will be Rs 1200.

2. The answer of your question no 2 will also be same as q.no. 1. we have first installment Rs100, then other installment will be 95, 90, 85..........and so on. as we know that common difference or c.d= difference between any two successive terms D= 90-95= -5 again we have Sn = n/2{2a+(n-1)d} 975=n/2{2x100+(n-1)x-5} after solving this equation you will have two values for n i.e. 15 and 26.

i have used "x" sign for multiplicatin in solution of you second querry. and please also send your feedback whether you are satisfied with the answer or not. And for also other querry of QA you can directly send your querry as private messege....... wish you all the best for your exams..........