I A man arrange to pay off a debt
o 3600 ley 40 annual instantemente
which foun an
an AP
A e
ucher
Gebes 30 intalnests
s
are paid, he dues baring one- cher
of the debt unpaid find the value
of the first installement ?
Answers
Answer:
\huge{\text{\underline{Correct\:Question:-}}}
CorrectQuestion:-
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:-}}}
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}}}⟹
1200
★ Total payment he paid in 30 installment:-
\implies⟹ S 30 = 3600 - 1200
\implies{\boxed{\tt{S 30 = 2400}}}⟹
S30=2400
S30 = 2400
★ Using formula:-
\large{\boxed{\text{Sn = n / 2 [2a + (n – 1) d]}}}
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}}}⟹
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}}}⟹
a = 51
Hence, the value of the first installment is 51
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