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Please solve this problem

CPT 5583 views 2 replies

1).a person pays Rs.975 by monthly instalment each less then the former by Rs.5.the first instalment is Rs.100.the time by which the entire amount will be paid is ?

ans:15months.

2).a person saved Rs.16500 in 10 years.in each year after the first year he saved Rs.100 more than he did in the preceeding year.the amount of money he saved in the 1st year was?

ans:Rs.1200.

3).if the sum of first n terms of an a.p is 2n^2+n then the difference of its tenth & first term is?

ans:36.

4).g is g.m.between a & b then 1/g^2-a^2+1/g^2-b^2=?

ans:1/g^2.

Replies (2)

For 1st Question:

 

As given in problem, A=100, D=-5


Formula is:

Sum of AP = n/2 [2a+(n-1)d], Therefore, 


975 = n/2 [200+(n-1)-5]

1950 = n(200-5n+5)

1950 = 205n-5n^2

5n^2-205n+1950 = 0

n^2-41n+391 = 0

n^2-26n-15n+390 = 0

n(n-26) - 15(n-26) = 0

Hence, n = 15 or 26.

So, n = 15 months.

 

Difference in savings of that man in two consecutive years is 100, which is constant. So, savings of that man are in AP.


As given in problem, n=10, S=16500, d=100

So, Formula is S=n/2 [2a+(n-1)d]

16500 = 10/2 [2a+(10-1)x100]

16500 = 5(2a+900)

2a = 3300-900 = 2400

Therefore, a=1200, which is his 1st year savings.


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