Question

A man arranges to pay off a debt of ₹3600 by 40 installments whifh form an AP series. When 30 of the installments are paid, he dies leaving one third of debt unpaid, find the value of the first installment!

Answer 1

$\boxed{Answer :}$

According to question we have given A.P. is

$\textbf{a, a + d, a + 2d + - - - - + 40th}$ (installment)

an = a + (n - 1) d

Then,

a40 = a + (40 - 1) d

a40 = a + 39d ........(A)

Now,

Sn = n/2 [a + an]

S40 = 40/2 [a + a + 39d] [from (A)]

S40 = 20 [2a + 39d]

Now, we have given that Total debt = Rs. 3600

So,

3600 = 20 [2a + 39d]

$\textbf{180 = 2a + 39d ......(1)}$

In the question it is said that one-third of the debt unpaid. Means,

Total debt ÷ 3 {it's one-third}

(3600 ÷ 3) = 1200

So,

Rs. 3600 - Rs. 1200 = Rs. 2400

Now, we have givrn S30 = 2400

S30 = 30/2 [a + a + 29d]

2400 = 15 [2a + 29d]

$\textbf{160 = 2a + 29d ......(2)}$

Solve (1) and (2) by elimination

+ 2a + 39d = + 180
+ 2a + 29d = + 160 (change the signs)
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+ 0a + 10d = + 20
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10d = 20

$\textbf{d = 2}$

Put value of d in (1)

2a + 39(2) = 180

2a + 78 = 180

2a = 102

$\textbf{a = 51}$

$\textbf{The value of first installment is}$

$\bf\huge{51}$
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Answer 2

Hii Mate

Plz refer to the attachment given..