Question

A man arrange to pay off a debt of rs.3600 by 40 annual instalment which are in
a.P. When 30 of the instalments paid he dies leaving one third of the dept unpaid. The value of 8th instalment is

Answer 1

$\huge{\text{\underline{Correct\:Question:-}}}$

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

$\huge{\text{\underline{Solution:-}}}$

★ Given:-

★ Total amount of Debt:-

$\implies$S 40 = 3600

★ Number of annual installments:-

$\implies$n = 40

★ Point to remember:-

He paid 30 installment and he dies leaving 1/3 of the debt unpaid.

Unpaid amount = ⅓ × 3600

$\implies{\boxed{\text{1200}}}$

★ Total payment he paid  in 30 installment:-

$\implies$S 30 = 3600 - 1200

$\implies{\boxed{\tt{S 30 = 2400}}}$

S30 = 2400

★ Using formula:-

$\large{\boxed{\text{Sn = n / 2 [2a + (n – 1) d]}}}$

★ For 30 installments:-

$\implies$S30 = 30 / 2 [2a + (30 - 1)d]

$\implies$2400 = 15 [2a + 29d]

$\implies$2400 / 15 =  [2a + 29d]

$\implies$160 = 2a + 29d

$\implies$2a = 160 - 29d

$\implies$2a + 29d = 160 $\implies$(1)

★ For 40 installments:-

$\implies$S40 = 40 / 2 [2a + (n - 1) d]

$\implies$3600 = 20 [2a + (40 -1) d]

$\implies$3600 / 20 = 2a + 39d

$\implies$180 = 2a + 39d

$\implies$2a + 39d  = 180 $\implies$ (2)

★ On subtracting eq (i) from (ii)

$\implies$ 10 d = 20

★ Therefore:-

$\implies$d = 20 / 10

$\implies{\boxed{\text{d = 2}}}$

On Putting the value of d = 2 in eq (1),

$\implies$2a + 29d = 160

$\implies$2a + 29 (2) = 160

$\implies$2a + 58 = 160

$\implies$2a = 160 - 58

$\implies$2a = 102

$\implies$a = 102 / 2

$\implies\large{\boxed{\text{a = 51}}}$

Hence, the value of the first installment is 51

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Answer 2

Question :-

A man arrange to pay off a debt of rs.3600 by 40 annual instalment which are in A.P. When 30 of the instalments paid he dies leaving one third of the dept unpaid. The value of 8th instalment is .?

Answer :-

→ The value of 8th instalment is 65

To find :-

Find the value of 8th instalment.

Formula used :-

$\boxed{s_n \: = \frac{n}{2} \big(2a + (n - 1)d \big)}$

Step - by - step explanation :-

Solution :-

Using the formula of sum of n terms of an Airthmatic progression.

• For 40 instalments = 3600

$\implies \: \small \: 3600 = \frac{40}{2} \big(2a \: + (40 - 1)d \big) \\ \\ \implies \: \frac{3600}{20} \: = 2a \: + 39 \: d \\ \\ \implies \: 180 = 2a + 39d \: \: \: .....(1)$

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$\bf{ \: unpaid \: amount \:} = \frac{1}{3} \times 3600 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1200$

Therefore ,

Total payment till 30 instalments

→ 3600 - 1200

→ 2400

• For 30 instalments = 2400

$\implies \: 2400 = \frac{30}{2} \big(2a \: + (30 - 1)d \big) \\ \\ \implies \: \frac{4800}{30} = 2a + 29 d \\ \\ \implies \: 160 = 2a + 29d \: \: \: ......(2)$

Now ,subtract eq.(2) from eq.(1)

Then we get,

$\implies \: 20 = 39d - 29d \\ \\ \implies \: 20 = 10d \\ \\ \implies \: \boxed{ d \: = 2} \\ \\ put \: d \: = 2 \: in \: eq.(2) \\ \\ \implies \: 160 = 2 a + 29 \times 2 \\ \\ \implies \: 160 - 58 = 2a \\ \\ \implies \: a \: = \frac{102}{2} \\ \\ \implies \: \boxed{ a \: = 51 }$

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Therefore ,

The value of 8th instalment is ↓

→ a+ (8-1) ×2

→ 51 + 14

→ 51+14

→ 65

Therefore,

The value of 8th instalment is 65

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