Question

only genius can answer this

a man arranges to pay a debt of ₹3600 by 40 annual installations which form an arithmetic series. When 30 installations were paid , he dies leaving one - third of the debt unpaid

value of its first installation is

(a) ₹102. (b) ₹560
(c) ₹51. (d) ₹52

Answer 1

Let the first installment be 'a' and common difference is 'd'.

Given, Sum of 40 installments is 3600.

We know that Sum of n terms of an AP sn = (n/2)[2a + (n - 1) * d]

⇒ 3600 = 40/2[2a + (40 - 1) * d]

⇒ 3600 = 20[2a + 39d]

⇒ 3600/20 = 2a + 39d

⇒ 180 = 2a + 39d    ----- (1)

Now,

Given, after 30 installments, he dies leaving one-third of the debt unpaid.

⇒ (1/3) * 3600

⇒ 1200.

So, the amount paid = 3600 - 1200 = 2400.

Hence, Sum of 30 installments is 2400.

We know that sum of n terms of an AP sn = n/2[2a + (n - 1) * d]

⇒ 2400 = 30/2[2a + (30 - 1) * d]

⇒ 2400 = 15[2a + 29 * d]

⇒ 160 = 2a + 29d    ------- (2)

On solving (1) & (2), we get

2a + 39d = 180

2a + 29d = 160

----------------------

       10d = 20

           d = 2.

Substitute d = 2 in (1), we get

⇒ 2a + 39d = 180

⇒ 2a + 39(2) = 180

⇒ 2a + 78 = 180

⇒ 2a = 102

⇒ a = 51.

Therefore, the value of first installment is 51.

Hope this helps!

Answer 2

Here your answer goes

Step :- 1

Let the first installment be a and Difference be d

Given ,

Sum of 40 installations = ₹3600

Step :- 2

Use the formula of Sn

$S_{n} = \frac{n}{2} ( 2a + ( n- 1 ) * d )$

Put the values

$3600 = 40/2 ( 2a + (40 - 1) * d )$

$\frac{3600}{20} = 2a + 39d$

180 = 2a + 39d ------->> 1

Step :- 3

Given , 1/3 of the debt unpaid When he dies and before  they were paid  30 installations

( 1/3 ) * 3600

==>  1200

Amount paid = 3600 - 1200

==> 2400

Therefore , the Sum of 30 installations is 2400

Step :- 4

 By Using the formula of Sn

$S_{n} = \frac{n}{2} ( 2a + ( n- 1 ) * d )$

Put the values in the given formula

==>  $2400 = 30/2(2a + (30 - 1) * d )$

==> $2400 = 15 ( 2a + 29 * d )$

==> 160 = 2a + 29d ------- >> 2

Step :- 5

As We have two variables  of linear pair We have 3 methods to obtain the result

Let us Solve by Substitution Method  because  it obtained  Solution easily and fast as Compare to other

On Solving  Equation 1 and 2 We get ,

$( 2a + 39d = 180 ) ( 2a + 29d = 160 ) --------->> 10d = 20$

d = 10/20

d = 2

Now ,

Substitute the value of d in equation ( 1 )

$2a + 39d = 180$

==> $2a + 39(2) = 180$

==> 2a + 78 = 180

==> 2a = 102

===> a = 102/2

====> a = 51

Therefore , The value of first installation is ₹51 ( Option :- C )

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