Question

a man agree to payoff a debt of 3600 by 40 annual installment which from in a. p. when 30 of the installments are paid he dies leaving 1/3 of the debt unpaid find the value of the first ​

Answer 1

Answer:

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Explanation:

MATHS

A man arranges to pay off a debt of RS. 3600 by 40 annual installments which form an arithmetic series. When 30 of the installments are paid, he dies leaving one-third of the debt unpaid, find the value of the first installment.

November 22, 2019avatar

Bipin Bhandari

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ANSWER

According to the question,

Tthe total amount of debit to be paid in 40 installment = Rs 3600

After 30 installment one-third of his debit is left unpaid. This means that he paid two third of the payment. So,

The amount he paid in 30 installments =

3

2

(3600)

= 2(1200)

= 2400

Let us take the first installment as a and common difference as d.

So, using the formula for the sum of n terms of an A.p,

s

n

=

2

n

[2a+(n−1)d]

Let us find a and d, for 30 installments.

s

30

=

2

30

[2a+(30−1)d]

2400=15[2a+(29)d]

15

2400

=2a+29d

160=2a+29d

a=

2

160−29d

.....(1)

Similarly, we find a and d for 40 installment.

s

40

=

2

40

[2a+(40−1)d]

3600=20(2a+(39)d)

20

3600

=2a+39d

a=

2

180−39d

subtracting (1) from (2), we get

a−a=(

2

180−39d

)−(

2

160−29d

)

0=

2

180−39d−160+29d

.....(2)

0=20−10d

Further solving for d

10d = 20

d=

10

20

d = 2 the value

subtracting the value of d in (1), we get,

a=

2

160−29(2)

=

2

160−58

=

2

102

= Rs 51

Answer 2

ANSWER

According to the question,

Tthe total amount of debit to be paid in 40 installment = Rs 3600

After 30 installment one-third of his debit is left unpaid. This means that he paid two third of the payment. So,

The amount he paid in 30 installments =

3

2

(3600)

= 2(1200)

= 2400

Let us take the first installment as a and common difference as d.

So, using the formula for the sum of n terms of an A.p,

s

n

=

2

n

[2a+(n−1)d]

Let us find a and d, for 30 installments.

s

30

=

2

30

[2a+(30−1)d]

2400=15[2a+(29)d]

15

2400

=2a+29d

160=2a+29d

a=

2

160−29d

.....(1)

Similarly, we find a and d for 40 installment.

s

40

=

2

40

[2a+(40−1)d]

3600=20(2a+(39)d)

20

3600

=2a+39d

a=

2

180−39d

subtracting (1) from (2), we get

a−a=(

2

180−39d

)−(

2

160−29d

)

0=

2

180−39d−160+29d

.....(2)

0=20−10d

Further solving for d

10d = 20

d=

10

20

d = 2 the value

subtracting the value of d in (1), we get,

a=

2

160−29(2)

=

2

160−58

=

2

102

= Rs 51

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